Question about Special Relativity and matching results

[boggledandnot]

Assume a light was emitted from the origins of two frames in relative motion.

Assume the light acquired some position (x,y,z,t) in an F frame. Then map that point using the lorentz transformations to (x',y',z',t') in a F' frame.

Can anyone prove under all conditions that the the light postulate in the F' frame agrees this lorentz transformations to (x',y',z',t') is the correct answer in its own frame?

 

[mfb]

I don't understand what exactly your question is, but there is a very general result that gives "yes" for all those questions: the Lorentz transformations form a group (called Lorentz group). No matter how you transform things, you will always get correct physics in all reference frames.

 

[boggledandnot]

I don't understand what exactly your question is, but there is a very general result that gives "yes" for all those questions: the Lorentz transformations form a group (called Lorentz group). No matter how you transform things, you will always get correct physics in all reference frames.

Yes, can you prove mathematically that the other frames agree with the any answer in the Lorentz group. That is what I am asking.

 

[pervect]

Assume a light was emitted from the origins of two frames in relative motion.

Assume the light acquired some position (x,y,z,t) in an F frame. Then map that point using the lorentz transformations to (x',y',z',t') in a F' frame.

Can anyone prove under all conditions that the the light postulate in the F' frame agrees this lorentz transformations to (x',y',z',t') is the correct answer in its own frame?

If by the light postulate you mean that the speed of light is constant, yes. On any lightcone originating at the coordinate origin in F, a constant speed of light implies that distance^2 = c^2 * time^2, or x^2 + y^2 + z^2 - c^2 t^2 = 0.

Now, x^2 + y^2 + z^2 - c^2t^2 is known as the Lorentz interval, and algebra will confirm that if you use a generalized boost, the Lorentz interval will not change as a result of applying the transform - it's invariant. This implies that if x^2 + y^2 + z^2 - c^2 t^2 = 0, then x'^2 + y'^2 + z'^2 - c^2 t'^2 = 0 in frame F'. Thus the Lorentz interval is independent of the frame of reference, and so is the speed of light.

One particularly easy to write proof. consider a standard Lorentz boost in the x direction. Using the concept of rapidity and hyperbolic functions http://en.wikipedia.org/wiki/Rapidity, we can write a Lorentz boost in the x direction as:

x' = x $\cosh \theta$ + c t $\sinh \theta$
c t' = c t $\cosh \theta$ + x $\sinh \theta$
(see the wiki article for details)

Then by direct substitution of the above, x'^2 - c^2 t'^2 = (cosh^2 $\theta$ - sinh^2 $\theta$) x^2 - c^2 t^2 (cosh^2 $\theta$ - sinh^2 $\theta$) , using the identity cosh^2 - sinh^2 = 1 we can simplify this to x^2 - c^2 t^2

It's not necessary to use rapidities if they are unfamiliar or intimidating, but using them makes the formal similarity between rotations leaving distances unchanged in 3d Euclidean space and Lorentz transforms leaving Lorentz intervals unchanged in 4d Minkowskii space.

 

[mfb]

Yes, can you prove mathematically that the other frames agree with the any answer in the Lorentz group. That is what I am asking.

That follows from the group structure. Showing that the group is well-formed is a typical homework question for students, I'm sure you can find the derivation in many textbooks.
Showing that electrodynamics and so on are Lorentz invariant is done in textbooks as well. And I guess you can find websites deriving it, too.

 

[boggledandnot]

Yes, but prove the primed frame agrees for all possible circumstances that the Lorentz transforms are correct.

If it is so simple prove it with the math.

 

[PeterDonis]

If it is so simple prove it with the math.

pervect showed you the math in post #3. You should be able to work from there on your own. We don't mind helping you to understand, but we are not going to post detailed proofs of results that are basic facts of relativity. That's what textbooks are for.

 

[Orodruin]

Apart from what has been said already, I think it should be mentioned that leaving the speed of light invariant is essentially the definition of the Lorentz group and you use this to derive the form of the transformations, not the other way around.

 

[boggledandnot]

pervect showed you the math in post #3. You should be able to work from there on your own. We don't mind helping you to understand, but we are not going to post detailed proofs of results that are basic facts of relativity. That's what textbooks are for.

Thanks for your advice on textbooks.

Anyway, after we have this mapped point using the math as I posted above, we get say (x',y',z',t') from (x,y,z,t) using the lorentz transformations which you call the math.

And I am not asking for detailed proofs of the lorentz transformations. That is not at all what I asked.

I asked once the point (x,y,z,t) maps to (x',y',z',t'), does the prime origin for example using the light postulate in its own frame agree (x',y',z',t') is the correct answer for the circumstances.

This is a simple enough request. This is not requesting a detailed proof. It is simple verification that (x',y',z',t') is the correct answer in the view of the primed frame. It is sort of like checking your work using long division by multiplying the result by the denominator to get to the dividend. Then you know the answer is correct.

That is what I am asking for here.

 

[boggledandnot]

Apart from what has been said already, I think it should be mentioned that leaving the speed of light invariant is essentially the definition of the Lorentz group and you use this to derive the form of the transformations, not the other way around.

Yes, the speed of light is invariant.

But, that is not what I am asking.

And, I know exactly how to derive x'^2 + y'2 + z'^2 = (ct')^2 from x^2 + y^2 + z^2 = (ct)^2 and vice-versa. It is trivial algebra.

 

[PeterDonis]

It is simple verification that (x',y',z',t') is the correct answer in the view of the primed frame.

The correct answer to what? If you mean the correct answer to where the light will end up in the primed frame, the answer is yes.

 

[boggledandnot]

The correct answer to what? If you mean the correct answer to where the light will end up in the primed frame, the answer is yes.

Once you have the answer (x',y',z',t') from (x,y,z,t) have you ever proved the primed origin agrees with this answer yes or no? I mean, if you are going to make claims about logic in the primed frame, then certainly it is science to request that the primed frame origin for example agrees this is the correct answer.

Can you demonstrate that (x',y',z',t') is the correct answer for the primed origin that applies the light postulate? As you know, whatever the calculation in the unprimed frame, the primed frame sees a spherical light wave. That calculation must simply agree with the primed origin's light postulate. It is a scientific double check of the lorentz answer.

The goal of the lorentz transformations is to perfectly match the view of the other frame no?

 

[PeterDonis]

Once you have the answer (x',y',z',t') from (x,y,z,t) have you ever proved the primed origin agrees with this answer yes or no?

I'm not understanding what sort of additional proof you need. Could you give more detail about what exactly you are looking for? For example, an outline of what a proof you would find acceptable would look like?

whatever the calculation in the unprimed frame, the primed frame sees a spherical light wave.

The light wave is spherical in both frames.

 

[Dale]

Can you demonstrate that (x',y',z',t') is the correct answer for the primed origin that applies the light postulate? As you know, whatever the calculation in the unprimed frame, the primed frame sees a spherical light wave.

Yes. This was done for you already in pervect's post 4 paragraph 2. What more do you want, or what did you not understand?

 

[boggledandnot]

I'm not understanding what sort of additional proof you need. Could you give more detail about what exactly you are looking for? For example, an outline of what a proof you would find acceptable would look like?
The light wave is spherical in both frames.

Sure, I would like to see a proof that the primed frame origin using the light postulate agrees with the unprimed frame's calculation using the lorentz transformations on the location of the light. So, the unprimed frame origin claims the light is located at a transformed (x'y'z't') under some conditions.

Now, given the conditions that made the unprimed origin conclude this, under those same conditions, does the primed origin agree that the light is located at (x'y'z't') using its light postulate under those same conditions?

Does such a proof exist?

You know, one set of circumstances should lead to (x',y',z',t') using the lorentz transformations for the unprimed frame and also (x',y',z',t') using the light postulate in the primed frame.

So, condensed

Unprimed origin
circumstances->light postulate (x,y,z,t)->LT (x',y',z',t')
Primed frame origin
circumstances->light postulate (x',y',z',t')

 

[mfb]

Yes it exists, and you'll find it in textbooks. See post #5 for example:

That follows from the group structure. Showing that the group is well-formed is a typical homework question for students, I'm sure you can find the derivation in many textbooks.
Showing that electrodynamics and so on are Lorentz invariant is done in textbooks as well. And I guess you can find websites deriving it, too.

 

[Ibix]

Define two frames S and S', where S' moves at velocity v in the +x direction. A pulse of light is emitted from the origin, moving at an angle θ to the x-axis, at the time when the origins of coordinates in the two frames coincide - t=t'=0.

Working in S, the coordinates of the light pulse at time t=T are $x=cT\cos\theta$ and $y=cT \sin \theta$. You can substitute those into the Lorentz transforms to get x', y' and t'. It's simple enough to then show that $x'^2+y'^2=ct'^2$ - in other words, that the transformed coordinates of the light pulse are exactly as far from the origin of the transformed coordinates as light could have traveled in the transformed time. I'll leave the algebra to you.

Was this what you were looking for? If so, you can get to it much more directly: you can derive the Lorentz transforms from the assumption that the speed of light is constant in all frames; therefore the Lorentz transforms must keep the speed of light constant.

 

[boggledandnot]

Yes it exists, and you'll find it in textbooks. See post #5 for example:

Really.

The group structure is based on the Minkowski metric that says, a light beam measures c in one frame iff that same light beam measures c in the other frame.

The metric does not prove that the transformation produces the same light beam that measures c in the other frame given a set of circumstances and its light postulate. It only produces a light beam that measure c period.

I am fully capable of proving the Minkowski metric, so that is not at all what I am asking. I assume you know how to do this no?

Let me simplify it.

Assume a light beam is emitted up the positive x-axis of both frames when their origins are common in the standard configuration.

Now, let's set the circumstances.
Assume the primed coordinate d' is now common with the unprimed origin.
Where does the unprimed origin conclude the light beam is in the coordinates of the primed frame.
Given d' and the unprimed origin are at the same place, where does the primed frame conclude the light beam is?

Are the the same. Note they must be.

 

[boggledandnot]

Define two frames S and S', where S' moves at velocity v in the +x direction. A pulse of light is emitted from the origin, moving at an angle θ to the x-axis, at the time when the origins of coordinates in the two frames coincide - t=t'=0.

Working in S, the coordinates of the light pulse at time t=T are $x=cT\cos\theta$ and $y=cT \sin \theta$. You can substitute those into the Lorentz transforms to get x', y' and t'. It's simple enough to then show that $x'^2+y'^2=ct'^2$ - in other words, that the transformed coordinates of the light pulse are exactly as far from the origin of the transformed coordinates as light could have traveled in the transformed time. I'll leave the algebra to you.

Was this what you were looking for? If so, you can get to it much more directly: you can derive the Lorentz transforms from the assumption that the speed of light is constant in all frames; therefore the Lorentz transforms must keep the speed of light constant.

Many thnks.
But, no that is not what I was looking for.

Please see my post below yours.

 

[Ibix]

Assume a light beam is emitted up the positive x-axis of both frames when their origins are common in the standard configuration.

Now, let's set the circumstances.
Assume the primed coordinate d' is now common with the unprimed origin.
Where does the unprimed origin conclude the light beam is in the coordinates of the primed frame.
Given d' and the unprimed origin are at the same place, where does the primed frame conclude the light beam is?

So the emission event is (x,t)=(-X,0) - i.e., to he left of the origin (in either coordinate system), with the beam moving to the right. The event you are calling "now" is (x',t')=(0,T') - i.e., at the origin of the primed frame some time after it was emitted.

Is this a correct description of your set up?

 

[boggledandnot]

So the emission event is (x,t)=(-X,0) - i.e., to he left of the origin (in either coordinate system), with the beam moving to the right. The event you are calling "now" is (x',t')=(0,T') - i.e., at the origin of the primed frame some time after it was emitted.

Is this a correct description of your set up?

The emission event is (0,0) for both frames. The clocks are synched to 0 when the light pulse is emitted and both origins are common.

The circumstances event for the unprimed frame is when the primed coordinate (d',0,0) is at its origin.

The circumstances event for the primed frame is when the unprimed origin (0,0,0) is at its coordinate (d',0,0).

 

[Ibix]

The emission event is (0,0) for both frames. The clocks are synched to 0 when the light pulse is emitted and both origins are common.

The circumstances event for the unprimed frame is when the primed coordinate (d',0,0) is at its origin.

The circumstances event for the primed frame is when the unprimed origin (0,0,0) is at its coordinate (d',0,0).

You haven't specified what the groups of three numbers and groups of two numbers are. I presume that your two-element group is (x,t) or (x',t') and your three element groups are (x,y,z) or (x',y',z'). On that assumption, there are two events.

The emission event is at (x,t)=(0,0), which is also (x',t')=(0,0).

The two "circumstances events" are two descriptions of the same event, which occurs at some unknown time when x=0 transforms to x'=d'.

Is this a correct description of your set up?

 

[boggledandnot]

You haven't specified what the groups of three numbers and groups of two numbers are. I presume that your two-element group is (x,t) or (x',t') and your three element groups are (x,y,z) or (x',y',z'). On that assumption, there are two events.

The emission event is at (x,t)=(0,0), which is also (x',t')=(0,0).

The two "circumstances events" are two descriptions of the same event, which occurs at some unknown time when x=0 transforms to x'=d'.

Is this a correct description of your set up?

I think you have it right. Sure looks like it.

Then, the 3rd event is the location of light along the positive x-axis in the coordinates of the primed frame when the circumstances event is true.

 

[Ibix]

Use the Lorentz transforms to solve for t and t' when x=0 -> x'=d'. Call those values T and T' respectively; you get T=-d'/vγ, and T'=-d'/v. Obviously, the x and x' coordinates of the light pulse at this event are cT and cT', respectively. So your third event is (x,t)=(cT,T), or (x',t')=(cT',T'). Note that this is before the emission event (assuming d' is positive) because of the way you specified your second event.

The "third event" is two separate events, because the two frames don't agree on what is "now"

I'm not quite sure where you are going with this.

 

[PeterDonis]

boggledandnot, your descriptions are not making things any clearer. I've read and re-read the statements quoted below and I still can't fit them all into a single coherent picture. See comments/questions below.

Unprimed origin
circumstances->light postulate (x,y,z,t)->LT (x',y',z',t')
Primed frame origin
circumstances->light postulate (x',y',z',t')

I suspect part of the problem is that you are using your own idiosyncratic terminology instead of standard terminology. For example, when you say "unprimed origin" or "primed origin", are you just referring to coordinates in the unprimed and primed frames, respectively? Or do you mean something else?

Assume the primed coordinate d' is now common with the unprimed origin.

What does this even mean?

when the primed coordinate (d',0,0) is at its origin.

What does this mean? It makes no sense.

when the unprimed origin (0,0,0) is at its coordinate (d',0,0).

This might make sense if it means "an object at rest in the unprimed frame at the spatial origin is at spatial coordinates (d', 0, 0) in the primed frame". But how does that fit in with anything else?

The usual way of specifying a scenario in SR is to give coordinates (x, y, z, t) of all events of interest in some chosen frame. Then you can see how they transform into some other frame using the Lorentz transformations. Can you give (x, y, z, t) coordinates for the events you're interested in? Even if you could give (x, y, z, t) for some and (x', y', z', t') for others, that would help. What you've given up to now isn't enough to understand what you're getting at. If we can't understand what you're asking, we can't answer.

 

[boggledandnot]

Use the Lorentz transforms to solve for t and t' when x=0 -> x'=d'. Call those values T and T' respectively; you get T=-d'/vγ, and T'=-d'/v. Obviously, the x and x' coordinates of the light pulse at this event are cT and cT', respectively. So your third event is (x,t)=(cT,T), or (x',t')=(cT',T'). Note that this is before the emission event (assuming d' is positive) because of the way you specified your second event.

I'm not quite sure where you are going with this.

OK, so let's make sure we are clear, at the circumstances event the correct answer for the location of the light beam in the coordinates of the primed frame is (cd'/v,0,0,d'/v) based on the light postulate yes or no. In other words, it is not at 2 different places.

 

[Ibix]

OK, so let's make sure we are clear, at the circumstances event the correct answer for the location of the light beam in the coordinates of the primed frame is (cd'/v,0,0,d'/v) based on the light postulate yes or no. In other words, it is not at 2 different places.

I messed up (it's after 2am here, so this will be my last post for now).

"Events" are positions and times. So the event you called the "circumstances event" is a position and a time. At the same time as that event, the light pulse is somewhere else. But because the two frames don't agree on what "at the same time" means (the relativity of simultaneity), the "third event" is actually two separate events. One is (x,t)=(-cd'/vγ,-d'/vγ) and the other is (x',t')=(-cd'/v,-d'/v). These are different places and times.

So, yes, the light is at two different places, but at two different times. Which is fine.

 

[harrylin]

[..] Can you demonstrate that (x',y',z',t') is the correct answer for the primed origin that applies the light postulate? As you know, whatever the calculation in the unprimed frame, the primed frame sees a spherical light wave. That calculation must simply agree with the primed origin's light postulate. It is a scientific double check of the lorentz answer. [..]

That's easy to do, and I think that others are already showing you how. Instead I'll answer the question that you didn't ask:
Probably the issue is related to relativity of simultaneity. If that is not set correctly, then the origin is placed wrongly.

And from the last post, I see that Ibix now also focuses on that point.

 

[pervect]

Sure, I would like to see a proof that the primed frame origin using the light postulate agrees with the unprimed frame's calculation using the lorentz transformations on the location of the light. So, the unprimed frame origin claims the light is located at a transformed (x'y'z't') under some conditions.

Now, given the conditions that made the unprimed origin conclude this, under those same conditions, does the primed origin agree that the light is located at (x'y'z't') using its light postulate under those same conditions?

Does such a proof exist?

I don't understand why you are unhappy with my proof - you didn't, for instance, ask any questions about it. I'm always interested in helping people understand, I'm personally not so interested in debate.

Be that as it may, if you have any questions about my post, feel free to ask. RL has been a bit hectic recently, but if time permits I'd be happy to explain further. There are many other people here who can help you to understand if if events prevent me from responding further.