## Miscellaneous Science Questions

For the discussion of the sciences. Physics problems, chemistry equations, biology weirdness, it all goes here.

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PM 2Ring
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### Re: RELATIVITY QUESTIONS! (and other common queries)

The main problem seems to be that I've been using the equation:

(delta)t' = (delta)t/(1-(u^2)/(c^2))

The time dilation formula isn't adequate for this problem, for the reason you've already mentioned. You need to use the Lorentz Transformation.

burningcowsrule
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Tried it with a Lorentz transformation, got a sensible looking answer. Thanks a lot

thefume79
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Plamo wrote:Relativity has been bothering me, mainly in the form of simultaneity, I'll take the example from one of Brian Greene's books:

Two guys are on a train, this train is fast moving (say, 7/8ths the speed of light) they are several meters apart.
They each have a stopwatch, which they agree to start when they see the light of a light bulb turn on.
The mediator on the train turns on the light, and, as relativity states, they start their watches at the same time from the perspective of the mediator.
The second mediator, standing on the train station that the train happens to pass by, however, disagrees, stating that the clocks were not started at the same time, as the light took longer to reach the guy on the front of the train, as it had to catch up to him at difference of 1/8th the speed of light, whereas the guy on the back approached the light at a combined speed of 15/8ths of the speed of light.

Now, I understand that simultaneity is relativistic, and each point of view has equal merit, and is considered true. For the first mediator, the clocks have the same time. For the second mediator, the clock at the front of the train will be behind than that of the back of the train, as I understand it.

My question is, what happens when the 1st mediator and his two clock bearing friends hop off the train, and meet the second mediator in the town between the two stations for some lunch and tea? Will the clocks be in synchronization, or will one be slower than the other?

My hypotheses are that either:
a) The deceleration of the train causes reverse time dilation, or something of the effect, causing the clock on the front of the train to 'catch up' to the other clock, resulting in the clocks to be simultaneous from the perspective of the stationary (No pun intended) observer
b) Greene(or I) is(am) missing some information or mechanic, or the experiment has been simplified to the point of inaccuracy

In any case, I'd love this experiment to be explained to me.

I read xkcd every other day, and somehow managed to google up this forum independently trying to find the answer similar to the question posed above

In The Elegant Universe, the example given is 2 heads of state are riding a train with a light bulb in the middle; the mediator flips the light on, and the 2 people (one in the front, one in the back) sign the treaty, from their perspective simultaneously. But observers off the train on a platform will see the light from the bulb reaching the guy on the back of the train before it reaches the guy on the front. I get all of that.

What I can't grasp is why would an observer on the platform see one guy signing before another? My brain is telling me the platform observer would see both sign at the same time, but just that one guy would be signing in the dark. What am I missing?

Thank you to whoever can help me understand.

phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Something to consider is that if two events happen at the same time and in the same place, then all observers will agree that they coincide. Like, if two cars crash, all observers will agree that the cars were trying to be at the same place at the same time, there won't be any observers that think that the collision just didn't happen. So, for instance, "the light reaches guy 1" and "guy 1 signs" are two events that happen at the same place at the same time... so everyone will agree they happen at the same time. We can't have one observer thinking the person signs before the light reaches them, and another observer thinking they sign after.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

After all, otherwise the only conclusion the man on the platform can make is that one head of state cheated.
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thefume79
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### Re: RELATIVITY QUESTIONS! (and other common queries)

phlip wrote:Something to consider is that if two events happen at the same time and in the same place, then all observers will agree that they coincide. Like, if two cars crash, all observers will agree that the cars were trying to be at the same place at the same time, there won't be any observers that think that the collision just didn't happen. So, for instance, "the light reaches guy 1" and "guy 1 signs" are two events that happen at the same place at the same time... so everyone will agree they happen at the same time. We can't have one observer thinking the person signs before the light reaches them, and another observer thinking they sign after.

OK, so what happens if it was a mechanical trigger (like 2 mediators walking from the center of the table and tapping the heads of state on the shoulder or something) instead of a light bulb? Would the platform guy still observe a difference in signing times (even if it was really small)? I just don't see what the light bulb has to do with anything. Is the train distorted from front to back? Or is it how the photons are reaching the platform? Does it matter where the platform is?

Sir_Elderberry wrote:After all, otherwise the only conclusion the man on the platform can make is that one head of state cheated.

Yeah I just can't get my head around 2 simultaneous physical events in 1 reference frame being viewed differently by a 3rd observer. But I see how it would have to be mathematically to keep c constant. I think. But basically if the front and back wall of the train had a clock, it would give the same to people within the car, but different times to the platform? Even if the platform viewed them from an equidistant?

and THANKS to both of you!

phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

thefume79 wrote:OK, so what happens if it was a mechanical trigger (like 2 mediators walking from the center of the table and tapping the heads of state on the shoulder or something) instead of a light bulb? Would the platform guy still observe a difference in signing times (even if it was really small)?

The result would be the same, but the maths would be more complicated if the two mediators were moving at slower-than-light speeds. You'd need to bring in the velocity-addition formula to find out what velocity the two mediators would be moving relative to the platform, and find out when they'd intersect the two ends of the train... but still, the one moving to the back of the train would arrive first, from the platform's POV.

thefume79 wrote:I just don't see what the light bulb has to do with anything.

Well, the "c is the same for everyone" rule makes it easier to work with light than other moving objects... for one, you don't have to mess around with adding velocities, because the velocity-addition formula is such that c + v = c for all v (which is basically the same thing as saying "c is the same for everyone").

thefume79 wrote:Is the train distorted from front to back?

Nope, the mediator is still in the middle of the train, as seen from the platform.

thefume79 wrote:Or is it how the photons are reaching the platform? Does it matter where the platform is?

Surprisingly, it doesn't matter where the platform is, the result is the same. And the travel time of the photons to the platform from the train is irrelevant... when we say "from the point of view of the platform, the event happened at time t", we mean after correcting for that travel time... like, if the train was a light second away, and the light arrived at us right now of an event, then we'd say that event happened one second ago (from our POV).

thefume79 wrote:But basically if the front and back wall of the train had a clock, it would give the same to people within the car, but different times to the platform? Even if the platform viewed them from an equidistant?

Yep. The one towards the back of the train would show a later time (ie it would reach, say, noon, before the one at the front of the train would).

thefume79 wrote:and THANKS to both of you!

No problem!

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bfollinprm
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### Re: Common Questions

Herman wrote:Inspired by the "Certainty" thread:

Quantum mechanics works for macroscopic ("everyday") stuff too. In fact it makes more accurate predictions than classical mechanics. We use classical mechanics because it's a good approximation and the math is easier. There is no sharply defined "quantum realm." The border between quantum and macroscopic phenomena is a convenient fiction and is subject to context, like the border between physics and chemistry.

This isn't necessarily true. We can't solve a quantum problem without the classical approximation for macroscopic systems, so we have no idea whether or not it makes a better fit (it could be worse). Our theories suggest that QM should work in macroscopic systems, but there isn't any experimental proof.

bfollinprm
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### Re: RELATIVITY QUESTIONS! (and other common queries)

doogly wrote:
Meteorswarm wrote:
gmalivuk wrote:
Meteorswarm wrote:conservation of energy seems to say you should have enough energy to overcome the black hole's pull, since you have your speed + whatever GPE you'd lose in approaching it. What gives?

Doesn't matter how much energy you have: if it's finite, it's not enough to overcome the black hole's pull.

But shouldn't you gain precisely as much energy as you'd need to get out by falling in in the first place?

That's an awfully Newtonian way of thinking, and it just doesn't apply here.

To expand: The energy isn't the problem, it's the curvature of space. In a black hole, space is curved sufficiently to switch space-time to what I call "time-space"--the spacial radial dimension and the time dimension are flipped. Skipping the math, you can't escape a black hole for the same reason you can't go backwards in time; the spacial "radial" direction only allows travel in one direction--inwards. It's the same as time in ordinary space-time, we can only travel in one direction--the future.

sikyon
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### Relativity Questions

So... how does inertia work in general relativity? Is inertia now a function of velocity like momentum is? Are they related?

If you go fast, does your gravity change?

Mostly I'm just curious, as I don't understand general relativity so well

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### Re: Relativity Questions

inertia is the resistence to acceleration that is associated with mass.

as mass increases as velocity does, so does inertia. This is the reason that your velocity cannot exceed the speed of light as an object exerting say 100N of force with a rest mass of 1Kg, the object starts accelerating at 100ms^-2 but by its mass then increases and the acceleration decreases.

mass is given by the equation

m0(1-V2/C2)-1/2 where m0 is the rest mass.
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sikyon
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### Re: Relativity Questions

eSOANEM wrote:inertia is the resistence to acceleration that is associated with mass.

as mass increases as velocity does, so does inertia. This is the reason that your velocity cannot exceed the speed of light as an object exerting say 100N of force with a rest mass of 1Kg, the object starts accelerating at 100ms^-2 but by its mass then increases and the acceleration decreases.

mass is given by the equation

m0(1-V2/C2)-1/2 where m0 is the rest mass.

Isn't it a momentum increase, not a mass increase?

I mean you could talk about relativistic mass, but is there any benefit in doing so?

Anyhow, so reading wikipedia, apparently inertia had to be redifined in general relativity. Do you know what this redefinition was... ie was inertia redefined in terms of what it is (resistance to acceleration) or was inertia just redefined in terms of the equations that express it?

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### Re: RELATIVITY QUESTIONS! (and other common queries)

I am referring to relativistic mass as this is the only mass that is linked to inertia. A better question than what is the benefit of talking about relativistic mass is what benefit is there to talking about rest mass? In the case of the ratio between force and acceleration (which is effectively inertia) relativistic mass is the relevant mass to use as a moving object exerts and so experiences a more powerful gravitational force.

I'm not entirely sure what the wikipedia article means when it talks about inertia as a geodesic deviation but the way I've always been taught of inertia in relativity is that it no longer refers to an object's resistance to acceleration but rather how closely it adheres to the path of a light beam on the same initial trajectory, alternatively how easily the object is moved by the curvature of spacetime and hence leads to my first relativistic definition.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

eSOANEM wrote:I am referring to relativistic mass as this is the only mass that is linked to inertia. A better question than what is the benefit of talking about relativistic mass is what benefit is there to talking about rest mass?

It's a frame independent quantity which means it will be fixed in any tensor equations that pop up. Since that's virtually any equation people care about, it's come to dominate discussion.

sikyon
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### Re: RELATIVITY QUESTIONS! (and other common queries)

So if I were to say... that something had no inertia, would that imply no rest mass? they could go faster than light?

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### Re: RELATIVITY QUESTIONS! (and other common queries)

sikyon wrote:So if I were to say... that something had no inertia, would that imply no rest mass? they could go faster than light?

Inertia is kinda more related to energy. No inertia (no resistance to change in velocity) means no energy, which makes it a little say it exists at all. So sure, you can have your thing with no energy go as fast as you want
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### Re: RELATIVITY QUESTIONS! (and other common queries)

sikyon wrote:So if I were to say... that something had no inertia, would that imply no rest mass? they could go faster than light?

no, something with no mass can only travel at the speed of light, like a photon. Something travelling faster than the speed of light would have imaginary mass

The graph of mass against V assuming rest mass of one and with C normalised to one we get this

this shows the fact that subluminal particles have real mass and superluminal particles imaginary mass very well but does not easily show that something travelling at the speed of light must have 0 mass.

Charlie! wrote:
sikyon wrote:So if I were to say... that something had no inertia, would that imply no rest mass? they could go faster than light?

Inertia is kinda more related to energy. No inertia (no resistance to change in velocity) means no energy, which makes it a little say it exists at all. So sure, you can have your thing with no energy go as fast as you want

a photon has 0 mass but does transmit energy so that isn't entirely true.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

eSOANEM wrote:
Charlie! wrote:
sikyon wrote:So if I were to say... that something had no inertia, would that imply no rest mass? they could go faster than light?

Inertia is kinda more related to energy. No inertia (no resistance to change in velocity) means no energy, which makes it a little say it exists at all. So sure, you can have your thing with no energy go as fast as you want

a photon has 0 mass but does transmit energy so that isn't entirely true.

I think I was going more along the lines of "a photon has mass 0 but it can't go at just any speed because it still has some inertia." That is, a photon in motion will remain in motion. In order to have truly arbitrary speed, i.e. no inertia, you need something to not only have 0 mass but also 0 energy.

I think.
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phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

A photon has zero rest mass, which implies that at any speed other than c, it would have zero energy. In order for it to carry energy (and hence, in a sense, to exist at all), it has to travel at the speed of light.

A photon travelling at the speed of light has energy, a relativistic mass, and inertia... because all three of those are basically the same thing.

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bfollinprm
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### Re: RELATIVITY QUESTIONS! (and other common queries)

eSOANEM wrote:I'm not entirely sure what the wikipedia article means when it talks about inertia as a geodesic deviation but the way I've always been taught of inertia in relativity is that it no longer refers to an object's resistance to acceleration but rather how closely it adheres to the path of a light beam on the same initial trajectory, alternatively how easily the object is moved by the curvature of spacetime and hence leads to my first relativistic definition.

geodesic == straight-line path through local space-time (equivalent to the path which light would take)

so, geodesic deviation is the difference between the trajectory of a beam of light and the trajectory of the particle under consideration, which is what you said.

Since in relativity inertial and gravitational mass are conceptually (as opposed to functionally) equivalent, you could also just say that inertia is the measure of how much an object distorts the local space-time metric.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Null geodesics are equivalent to the path light would take. Timelike geodesics are for freefalling observers, and spacelike geodesics also exist and have geometrical interest.

http://en.wikipedia.org/wiki/Geodesic_d ... n_equation is something quite unlike inertia, you have to consider more than one geodesic for it to make sense.
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I think people here have an antiquated conception of inertia. In classical mechanics, m is called inertia because it is inversely proportional to acceleration, by the simple equation: $\textbf{F} = \frac{d \textbf{p}}{dt} = \frac{d}{dt} m \textbf{v} = m \frac{d \textbf{v}}{dt} = m \textbf{a}$ Clearly, this does not apply in SR, because momentum is not mv, but $\textbf{p} = γm\textbf{v} = \frac{m\textbf{v}}{\sqrt{1-\frac{v^2}{c^2}}}$ Even if we define [imath]γm = m_{rel}[/imath], that doesn't help much because $\textbf{F} = \frac{d \textbf{p}}{dt} = γ^3 m \textbf{a} \ne m_{rel} \textbf{a}$ So clearly mrel isn't really "inertia" either. Since acceleration depends not only on force and mass, but also on speed, no one quantity adequately describes inertia in SR as it did in classical mechanics.

Of course, that doesn't mean we can't come up with new conceptions of what "inertia" means, but if we do that, why wouldn't we just use rest mass? And as it is, "mass" is almost universally understood to mean "rest mass," and the term "relativistic mass" is essentially deprecated.

EDIT: There is one instance where using m to mean relativistic mass is nice. If m is relativistic mass, [imath]E = m c^2[/imath] is mechanical energy, not just rest energy. Otherwise, mechanical energy [imath]E = γ m c^2[/imath], not [imath]m c^2[/imath], as is commonly believed.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

hey people, I have a question about observable universe horizon - can things cross it in/out in finite time? for example, would light of source crossing out be basically red-shifted out of existence? is there existing accurate analysis of this somewhere, or do I have to actually take LCDM equations and solve it myself (not really trust myself to do this)?

Twistar
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### Special Relativity, Velocity additions and mass energy

First of all, I don't know if this should actually go in the generic relativity questions thread. I'm asking about some specific math and a derivation so I feel like this kind of justifies its own thread.

So we learned the basics of special relativity in my physics class this week. Its actually a physics I class so normally we wouldn't learn this but our professor was out for the week and decided to have the substitute teach this to us using his notes. The problem is with all of this stuff I think some stuff got miscommunicated and we learned some stuff a little bit off. However, my understanding is pretty shaky so I wanted to ask you guys.
This first problem I have is with velocity transformations for the perpendicular component of velocity. So for when the object has a horizontal and vertical component of velocity in one reference frame and then we go into another reference frame that is moving in the horizontal direction as compared to the first. There will clearly be the velocity additions in the horizontal direction but there will also be a transformation for the vertical component of velocity. We learned the formula like this:
$u_y=\frac{y}{t}$
$u_y=\frac{y'}{\gamma t'}$
$u_y=\frac{u_y'}{\gamma}$
my problem is with the second step. Basically we said the person in the non-moving reference frame experience time dilation that the other person does not so a gamma factor is introduced. The first problem my friend brought up with this is that since gamma with this formula we could introduce a reference frame moving with a very high v that could cause gamma to become very large and thus force [imath]u_y'[/imath] to become greater than c. instead I think we should have used [imath]t=\gamma(t'+\frac{v}{c^2}x')[/imath] to get
$u_y=\frac{u_y'}{\gamma(1+\frac{v}{c^2}u_x')}$
I actually cheated and looked this up and this is the equation I found for the perpendicular velocity transformation. So, I guess I'm first just wondering if all of my suspicions here are correct because the sub later went on to use the previous formula to prove mass energy equivalence. I guess I'll explain the proof.

Take two particles that are going to collide. They each have horizontal components of velocity a (pointing towards eachother) and vertical components of velocity b pointing towards eachother. After the collision the particles maintain their horizontal velocities and reverse their vertical velocities (particle A starts top left, particle B starts bottom right). I believe this is a conceivable collision due to the symmetry. Then we moved into the reference frame in which particle A has no horizontal velocity. We then said particle A has a velocity downward of u and B has a velocity upwards of [imath]\frac{u}{\gamma}[/imath] and that particle B has some velocity V to the left. the vertical components reverse for the post-collision conditions. we then did a balance of momentum (where [imath]p=m(v)v[/imath] to derive the equation for m(v) and went on from there. We took the case where the vertical component of velocity goes to 0 to find the relation. The problem is that it relied on the fact that the two horizontal components of velocity differed by a factor of [imath]\gamma[/imath] and if I use the latter velocity formula I stated I get a very different relation.

My questions are, Am I reading this all correctly or am I missing anything? If not is this is valid scenario to consider to derive mass energy equivalance using momentum? If it would help I can show you the math I got with all of the a's and b's still in the equations.
Thank you guys for your help. Let me know if there's any way I can help you understand anything I might have explained poorly.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

Eebster the Great wrote:I think people here have an antiquated conception of inertia. In classical mechanics, m is called inertia because it is inversely proportional to acceleration, by the simple equation: $\textbf{F} = \frac{d \textbf{p}}{dt} = \frac{d}{dt} m \textbf{v} = m \frac{d \textbf{v}}{dt} = m \textbf{a}$ Clearly, this does not apply in SR, because momentum is not mv, but $\textbf{p} = γm\textbf{v} = \frac{m\textbf{v}}{\sqrt{1-\frac{v^2}{c^2}}}$ Even if we define [imath]γm = m_{rel}[/imath], that doesn't help much because $\textbf{F} = \frac{d \textbf{p}}{dt} = γ^3 m \textbf{a} \ne m_{rel} \textbf{a}$ So clearly mrel isn't really "inertia" either. Since acceleration depends not only on force and mass, but also on speed, no one quantity adequately describes inertia in SR as it did in classical mechanics.

Of course, that doesn't mean we can't come up with new conceptions of what "inertia" means, but if we do that, why wouldn't we just use rest mass? And as it is, "mass" is almost universally understood to mean "rest mass," and the term "relativistic mass" is essentially deprecated.

We don't need a new conception of inertia for SR. The old one works just fine. Its the conceptions of force, velocity, and acceleration that have to change. F = dp/dt = ma works just fine in SR as long as F, p, and a are all four-vectors.

Twistar
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Dear mod, thank you for assimilating my thread. Sorry I got it in the wrong place.

Also, for my problem the professor somehow got the right stuff out of it, some of the math was just dubious to me I guess but like I said it might be me thats wrong. Thanks again

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### Re: RELATIVITY QUESTIONS! (and other common queries)

GMontag wrote:We don't need a new conception of inertia for SR. The old one works just fine. Its the conceptions of force, velocity, and acceleration that have to change. F = dp/dt = ma works just fine in SR as long as F, p, and a are all four-vectors.

While it is true that Fμ = mAμ, where F is the four-force, m is the invariant mass, and A is the four-acceleration, this does not give a complete picture, as the four-acceleration is the time derivative of the proper velocity (with a null time component), not the time derivative of the ordinary velocity. The key difference is that a, the ordinary acceleration, is what we would traditionally associate with classical notions of acceleration, while A, the four-acceleration, has very different properties.

But you're missing my point. I said "Of course, that doesn't mean we can't come up with new conceptions of what "inertia" means, but if we do that, why wouldn't we just use rest mass?" That is essentially what you are doing (you argue it is the notion of force and acceleration that change, not of inertia, but since inertia is essentially defined based on force and acceleration, if they change, it does too). You have changed the notion of inertia, but are still using rest mass, not relativistic mass. Hence, rest mass is the more useful term here.

Notably, this means a photon has no inertia, and thus objects do not need inertia to have momentum (and thus energy). This is also far more reasonable in GR where inertial mass is taken to be the same as passive gravitational mass, and thus only rest mass makes sense.

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### Re: RELATIVITY QUESTIONS! (and other common queries)

makc wrote:hey people, I have a question about observable universe horizon - can things cross it in/out in finite time? for example, would light of source crossing out be basically red-shifted out of existence? is there existing accurate analysis of this somewhere, or do I have to actually take LCDM equations and solve it myself (not really trust myself to do this)?

that depends on whether the universal acceleration in speeding up, slowing down, or constant.

If the universal acceleration is constant, then the observable universe horizon is a stationary event horizon. Things can cross out, but not in. Because outside that point the distance between us and a given object is increasing faster than the speed of light. And since nothing can move faster than light, nothing can travel TOWARDS us fast enough to come into visible range.

If the universal expansion is slowing down then the visible event horizon is getting bigger (assuming it eventually goes to 0 we would be able to see the entire universe given enough time), and things can pass through it in both directions. If The universal expantion is accelerating, then not only is the visible horizon an event horizon as per the constant expansion, but it's actively getting smaller. In fact this is one of the theoretical ends of the universe, because all the fundamental forces (as with everything else) are bound by the speed of light. And as the expansion speeds up a smaller and smaller volume is within our visible horizon. When that horizon shrinks so that the earth can't see the sun, the light will go out. And when it's smaller than the earth, the earth will shatter and each individual have their own visible event horizon, people further away disappearing until finally it's so small that your toes and head aren't connected and finally you're ripped apart at a subatomic level.
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Fat Tony
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### Re: RELATIVITY QUESTIONS! (and other common queries)

According to the Law of Conservation of Energy, a gun firing a bullet should cause the gun to have a momentum equal in magnitude to that of the bullet. Why, then, can a human shoulder easily stop a recoiling rifle with no penetration of the skin much more easily than it could a bullet?
Edit: Nevermind, I think I figured it out. The butt of the rifle distributes the force across a larger surface, lowering the pressure. Something along those lines?
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phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Well, for one, a rifle is physically bigger than the point of a bullet, so the pressure is significantly less - the force gets spread out more. For two, when you fire the gun, the bullet is accelerating for the whole length of the barrel (as opposed to when the bullet impacts, and, if it wasn't going to penetrate the skin, would have to stop in a much shorter distance)... so the force gets spread out over time, too. Plus, when you're firing a gun, you're expecting it, and can brace yourself to absorb the recoil, and have the opportunity to align the gun with the appropriate load-bearing parts of the shoulder... none of which is an option when you're being shot.

If a gun was made where the shoulder brace was filed down to a point, and it didn't have a barrel, just accelerated the bullets to superfast instantly, you rammed the point-brace wherever, and fired it without warning... it'd probably do just as much damage as the bullet that was fired.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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thoughtfully
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Another thing: according to Conservation of Momentum, the sum of the momenta of the gun and the bullet will be the same before and after the shot. Momentum is mass times velocity. The mass of the gun (or gun plus the shooter's body, if the gun is well braced) is much greater than the mass of the bullet, so the change in velocity is much smaller.

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Mr_Rose
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### Re: RELATIVITY QUESTIONS! (and other common queries)

Fat Tony wrote:According to the Law of Conservation of Energy, a gun firing a bullet should cause the gun to have a momentum equal in magnitude to that of the bullet. Why, then, can a human shoulder easily stop a recoiling rifle with no penetration of the skin much more easily than it could a bullet?
Edit: Nevermind, I think I figured it out. The butt of the rifle distributes the force across a larger surface, lowering the pressure. Something along those lines?

You also seem to have mixed up conservation of energy and conservation of momentum; they are conserved separately and both of them must be considered across the whole system. Also, if the gun is braced properly, theoretically the firer's body becomes part of the gun/bullet system. Not to mention that larger rifles have recoil dampening springs that spread out the impulse of the shot and improve both aim and wear on the rifleman. Oh and even smaller rifles tend to be semi-automatic these days, in which case part of the energy of the shot is spent loading the next round as well.
Microevolution is a term — when used by creationists — that is the evolutionary equivalent of the belief that the mechanism you use to walk from your bedroom to the kitchen is insufficient to get you from New York to Los Angeles.

phlip
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### Re: RELATIVITY QUESTIONS! (and other common queries)

thoughtfully wrote:Another thing: according to Conservation of Momentum, the sum of the momenta of the gun and the bullet will be the same before and after the shot. Momentum is mass times velocity. The mass of the gun (or gun plus the shooter's body, if the gun is well braced) is much greater than the mass of the bullet, so the change in velocity is much smaller.

Sure, but even if you, say, doubled the mass of the gun, when you fired it, the recoil would have half the velocity... but then to resist that recoil, your shoulder has to apply twice as much force, to counter the gun's inertia. It comes out the same, in the end. In an idealised system (no friction, no other forces on the gun except the propellant and your shoulder), the force the gun applies to your shoulder is equal to the force the gun applies to the bullet.

Actually, I suppose if your shoulder starts moving backwards as a result of the gun firing, giving you time to absorb the recoil... if the gun is heavier and slower, then your shoulder won't have to move as fast, so it can take longer before reaching the limit of its movement... so even though the momentum is the same, the time you have to absorb it is longer, so F=p/t is smaller.

Also, the mass can have other indirect effects, like if it's resting on the ground, then it can change how much of the recoil is absorbed by friction, for instance.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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other
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I think this is a relativity question, and I don't think it has already been covered in this thread, so I hope this is the correct place to post it:

1) a single photon with high energy can decay into a positron/electron pair, yes?

2) a single photon with low energy cannot experience such a decay, if the energy content of the photon is less than the sum of the rest masses

but ...

3) the energy content of a photon depends on the reference frame in which the measurement is made, so two reference frames can be designed such that observer A measures the photon as having sufficient energy to decay, while observer B does not. In such a pair of frames, if A actually does observe a decay in this mode, what does B observe? do the set-up requirements prevent A from telling B that the decay occured and then recieving a reply?

Caveats:

-Physics was a long time ago. I may be asking a question a question as silly as "Why is Blue heavy?" ( in which case the answer is probably "learn more physics!", but specific pointers would be helpful).

-I may have mis-remembered the posibility of single-photon decay, since I am partly extrapolating fom CPT reversal of electron/positron anihilation being able to produce exactly one photon if they have matched spin states (that is: Spin 1/2, charge -1, lepton number 1 + Spin 1/2, charge +1, lepton number -1 => Spin 1, charge 0, leptons 0; which can be satisfied by any odd number of photons having a correct net momentum, including a single photon). If the photon must interact with something in order to decay, then A & B may agree that the decay occured, but disagree as to which of the interacting entities provided the nessesary energy to allow the rest mass to appear.

- since the problem as stated involves both quantum and relativistic effects, the answer may depend on "which flavor of GUT do you subscribe to?" (and I probably won't understand any of the math that comes with the answers)

I came up with this question while considering something else ... whether a single photon with sufficient energy can form a black hole. I found a thread that touches on that, but I did not entirely comprehend the answer (uh ... "Does light have Gravity", I think was the thread title; unfortunately I've navigated away from it so cannot immediately post a link). I think this question isolates one of the things I was confused about, which is if there are discontinuities in the behavior of a photon at different measured energy levels, what happens when a photon has energy levels on either side of the discontinuity in different reference frames?

Thanks,
Other

crazyjimbo
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### Re: RELATIVITY QUESTIONS! (and other common queries)

other wrote:
3) the energy content of a photon depends on the reference frame in which the measurement is made, so two reference frames can be designed such that observer A measures the photon as having sufficient energy to decay, while observer B does not. In such a pair of frames, if A actually does observe a decay in this mode, what does B observe? do the set-up requirements prevent A from telling B that the decay occured and then recieving a reply?

The rest frame makes things simple since we only need to deal with the rest masses of the produced particles. If we move to another frame then the produced particles will need to be created with additional energy since they will be be moving in the new frame. This energy will be the same as that gained by the photon by moving to the new frame.

So physics still works... phew. I'm sure one day someone will ask a question like this and everyone will just go 'well f***... it's broken'

Mr_Rose
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### Re: RELATIVITY QUESTIONS! (and other common queries)

crazyjimbo wrote:So physics still works... phew. I'm sure one day someone will ask a question like this and everyone will just go 'well f***... it's broken'

Yah, but that will be the greatest day in all of history because we will discover so many new things, like maybe a way to go FTL without cheating.
Microevolution is a term — when used by creationists — that is the evolutionary equivalent of the belief that the mechanism you use to walk from your bedroom to the kitchen is insufficient to get you from New York to Los Angeles.

other
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### Re: RELATIVITY QUESTIONS! (and other common queries)

crazyjimbo wrote:
other wrote:
3) the energy content of a photon depends on the reference frame in which the measurement is made, so two reference frames can be designed such that observer A measures the photon as having sufficient energy to decay, while observer B does not. In such a pair of frames, if A actually does observe a decay in this mode, what does B observe? do the set-up requirements prevent A from telling B that the decay occured and then recieving a reply?

The rest frame makes things simple since we only need to deal with the rest masses of the produced particles. If we move to another frame then the produced particles will need to be created with additional energy since they will be be moving in the new frame. This energy will be the same as that gained by the photon by moving to the new frame.

So physics still works... phew. I'm sure one day someone will ask a question like this and everyone will just go 'well f***... it's broken'

Saddly, I think this makes things worse rather than better. I'll try to spell it out in more detail:

1) since I don't have the nessesary reference books handy, I'll call the rest mass of a positron-electron pair T (for "threshold")

2) Observer A measures a photon has having energy T + epsilon, moving in the positive-x direction (and, obviously, at c)

3) convienently, at time zero, the photon decays into a positron-electon pair, right in front of A at the co-ordinate x=0, y=0. In order to preserve momentum, the pair are separating slowly in the y- direction, and both have a small velocity component in the positive-x direction. the exact velocities and the angle between them depend on epsilon. This defines the "event" of interest; I'm not sure if A's measurement of the photon energy is also an event we need to consider.

4) B is moving parallel to the photon (that is, in the positive-x direction), at v = 0.9999 c. I fairly sure this means that he observes the same photon at an energy lower than (T + epsilon); I am not certain how to calculate the reduction factor which I believe is normally denoted by lambda. In order to preserve the question, (T + epsilon)/lambda must be less than T.

5) assuming that at time t=0, B is also at x =0, y=0, what does he observe? If he does see the decay, not only has a photon with energy less than T decayed into a collection of particles with rest mass T, they are also receding from him at velocities a shade under 0.9999c, so they also have considerable Kinetic energy!

douglasm
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### Re: RELATIVITY QUESTIONS! (and other common queries)

I'm fairly certain the answer to this dilemma is that a single photon by itself cannot decay into particles. You need additional photons or particles for it to interact with in order for decay to be possible. Then, once you find a configuration where decay actually is possible, you will find that any change in reference frame that reduces the energy of one photon or particle will give a corresponding decrease in the energy of produced particles and increase in the energy of other pre-decay photons or particles.

PM 2Ring
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### Re: RELATIVITY QUESTIONS! (and other common queries)

As mentioned, a photon in free space can't decay into a electron-positron pair. A high energy photon passing close to a nucleus can induce pair production, since it's interacting with the magnetic field of the nucleus. The nucleus is also involved in momentum conservation. (On a related note, a photon can't interact with a free electron.)

Going in the other direction, a electron-positron pair will decay into a minimum of two photons; the number will be odd or even depending on whether the lepton spins are parallel or anti-parallel. See the Wikipedia article on Positronium for details.

other
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### Re: RELATIVITY QUESTIONS! (and other common queries)

PM 2Ring wrote:As mentioned, a photon in free space can't decay into a electron-positron pair. A high energy photon passing close to a nucleus can induce pair production, since it's interacting with the magnetic field of the nucleus. The nucleus is also involved in momentum conservation. (On a related note, a photon can't interact with a free electron.)

Going in the other direction, a electron-positron pair will decay into a minimum of two photons; the number will be odd or even depending on whether the lepton spins are parallel or anti-parallel. See the Wikipedia article on Positronium for details.