How to explain Einstein's Special theory of Relativity.


by Lochlan.H
Tags: explanation, relativity
tensor33
tensor33 is offline
#73
Dec12-12, 12:49 AM
P: 52
Quote Quote by John Huang View Post
Yes, Einstein claimed that he proved LT so that you can say "at the beginning" SR did include LT but then SR claimed {Δt' = Δt/γ is the equation for the relation of the speed of time in two systems}. I think, after that, SR cannot claim to include LT any more.

In LT, the relation of the speed of time in two systems is not so simple, it is Δt' = γ(Δt-(vΔx/c^2)) and it is different from Δt' = Δt/γ except when x' is a constant.
For the last time, LT is a part of SR
In special relativity (or, hypothetically far from all gravitational mass), clocks that are moving with respect to an inertial system of observation are measured to be running more slowly. This effect is described precisely by the Lorentz transformation.
.
And you were right about Δt' = γ(Δt-(vΔx/c^2)) but wrong about Δt' = Δt/γ. When Δx=0, Δt'=γΔt not Δt/γ. I don't understand where you got Δt' = Δt/γ. I've never seen this equation used and it seems to be what is getting in your way.
JesseM
JesseM is offline
#74
Dec12-12, 01:05 AM
Sci Advisor
P: 8,470
Quote Quote by John Huang View Post
but then SR claimed {Δt' = Δt/γ is the equation for the relation of the speed of time in two systems}
Like I said before, Einstein didn't say anything about the "speed of time", that equation is intended to tell you how much time Δt elapses in the unprimed frame between two readings on a clock, if the clock itself shows an elapsed time of Δt' between those readings (so Δt' is also the time in the clock's rest frame, where Δx'=0). For example, if the readings are "10 seconds" and "15 seconds", then Δt' = 5 seconds, and in the second frame the clock is running slow so in this frame it will take a longer period of Δt = γ*5 seconds for the clock's second hand to move from 10 to 15.

Look at section 4 of Einstein's 1905 paper, where he uses t for the time elapsed on the clock in the "stationary system" and τ for the time elapsed in the "moving system" when the clock itself is at rest relative to the "moving system", i.e. the time between the two readings in clock's own rest frame, introducing the setup like this:
Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time τ when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time τ. What is the rate of this clock, when viewed from the stationary system?
He then goes on to show that the equation τ = t/γ describes the relation between these two times in this scenario, using the LT to derive it.

So, after having looking at that section of the 1905 paper, would you agree that Einstein was not talking about the "speed of time" but rather meant the equation Δt' = Δt/γ to apply to the time elapsed between two readings of a physical clock, as measured both in its own rest frame (where Δx'=0) and in the frame where it's moving? Would you agree that given those assumptions about what the equation is supposed to represent, Δt' = Δt/γ can be derived directly from the LT? It's easy to show why, since one of the LT equations is Δt = γ(Δt' + vΔx'/c2), so if you set Δx'=0 you are left with Δt = γΔt' or Δt' = Δt/γ.

If you disagree with any of this, please explain why you think I am mistaken about what Einstein meant, preferably pointing to a specific quote from the paper to make your case.
JesseM
JesseM is offline
#75
Dec12-12, 01:09 AM
Sci Advisor
P: 8,470
Quote Quote by tensor33 View Post
And you were right about Δt' = γ(Δt-(vΔx/c^2)) but wrong about Δt' = Δt/γ. When Δx=0, Δt=γΔt not Δt/γ. I don't understand where you got Δt' = Δt/γ. I've never seen this equation used and it seems to be what is getting in your way.
It's an arbitrary matter of convention which frame to define as the clock's own rest frame, although most sources do use unprimed for the frame where the clock is at rest, you can find the occasional source that defines unprimed as the observer's frame like this one from a university website, it's not really "wrong" as long as you explain which frame is which.
tensor33
tensor33 is offline
#76
Dec12-12, 01:15 AM
P: 52
Quote Quote by John Huang View Post
Yes, Einstein claimed that he proved LT so that you can say "at the beginning" SR did include LT but then SR claimed {Δt' = Δt/γ is the equation for the relation of the speed of time in two systems}. I think, after that, SR cannot claim to include LT any more.

In LT, the relation of the speed of time in two systems is not so simple, it is Δt' = γ(Δt-(vΔx/c^2)) and it is different from Δt' = Δt/γ except when x' is a constant.
I see your mistake. You are saying Δt' = Δt/γ when the equation is Δt=Δt'/γ. There is no contradiction between Δt=Δt'/γ and Δt'=γΔt. They are simply inverses. Someone correct me if I'm wrong.
tensor33
tensor33 is offline
#77
Dec12-12, 01:22 AM
P: 52
Quote Quote by JesseM View Post
It's an arbitrary matter of convention which frame to define as the clock's own rest frame, although most sources do use unprimed for the frame where the clock is at rest, you can find the occasional source that defines unprimed as the observer's frame like this one from a university website, it's not really "wrong" as long as you explain which frame is which.
But if we do define the unprimed frame as the one where the clock is at rest, wouldn't the equation Δt' = Δt/γ be incorrect?
JesseM
JesseM is offline
#78
Dec12-12, 01:24 AM
Sci Advisor
P: 8,470
Quote Quote by tensor33 View Post
I see your mistake. You are saying Δt' = Δt/γ when the equation is Δt=Δt'/γ. There is no contradiction between Δt=Δt'/γ and Δt'=γΔt. They are simply inverses. Someone correct me if I'm wrong.
I would say it's wrong that "the" time dilation is Δt=Δt'/γ rather than Δt' = Δt/γ, it all depends on which frame is chosen to be the one where the clock is at rest. If the clock is at rest in the unprimed frame the first is correct, if at rest in the primed frame the second is correct.
JesseM
JesseM is offline
#79
Dec12-12, 01:25 AM
Sci Advisor
P: 8,470
Quote Quote by tensor33 View Post
But if we do define the unprimed frame as the one where the clock is at rest, wouldn't the equation Δt' = Δt/γ be incorrect?
Yes, in that case it would be incorrect, but John Huang hasn't specified that either is supposed to be the rest frame of a specific clock, and doesn't seem to understand that the time dilation equation is meant to deal specifically with the case of a clock at rest in one of the two frames (hopefully he will read my most recent comment to him).
harrylin
harrylin is offline
#80
Dec12-12, 03:22 AM
P: 3,178
Quote Quote by John Huang View Post
Yes, Einstein claimed that he proved LT so that you can say "at the beginning" SR did include LT but then SR claimed {Δt' = Δt/γ is the equation for the relation of the speed of time in two systems}. I think, after that, SR cannot claim to include LT any more.

In LT, the relation of the speed of time in two systems is not so simple, it is Δt' = γ(Δt-(vΔx/c^2)) and it is different from Δt' = Δt/γ except when x' is a constant.
SR claims the different solutions for x' is a constant and for x is a constant, as I summarized in my post #64. In words those equations boil down to what Dalespam wrote in his post #55.

Do you remember where you read that wrong explanation of SR that you keep hanging on?
ghwellsjr
ghwellsjr is online now
#81
Dec12-12, 04:12 AM
PF Gold
P: 4,537
Quote Quote by John Huang View Post
My point is a logical issue.

In above example, two systems have constant relative velocity so that the speed of time in the moving system t' and the speed of time in the stationary system t should be decided once we select the point O as the stationary point, and the O' as the moving point. Under this SPECIFIC arrangement, when we talk about a period of time for ONE SPECIFIC EVENT then we should have ONLY ONE event period Δt as recorded in the stationary system and ONLY ONE event period Δt' as recorded in the moving system.

Now, what SR claims is Δt' = Δt/γ and what LT claims is Δt' = γΔt for the ABOVE example. Logically speaking, this should not happen UNLESS γ=1, isn't it? How do you explain this logical issue?
...
You are correct that LT claims Δt' = γΔt when x=0. Here is a graph to illustrate a clock at rest at x=0. The blue dots represent ticks of the clock that occur every second:



Now we use the LT to transform the coordinates of all the events (the 1-second ticks represented by blue dots) into the coordinates of another frame moving at 0.866c with respect to the first frame. When we do the LT, we always call the original frame the unprimed frame and the new frame the primed frame:



Note that since the speed is 0.866c, gamma is equal to two. You can see that the first event which occurred at the time coordinate of 1 in the first frame occurs now at the time coordinate of 2 in the second frame, just like the LT formula indicates.

So far so good.

Now we want to see how to use the time dilation formula. Unlike the LT formula which applies to the coordinate times in two different frames, the time dilation formula applies to the relationship between the time (primed) on a clock compared to the coordinate time (unprimed), all in a single frame.

So let's apply the time dilation formula to the first graph where the speed of the clock is zero and gamma equals one. The formula says that the time on the clock is the same as the coordinate time and we see that because the spacing of the dots is the same as the spacing of the grid lines for the graph.

Now let's apply the time dilation formula to the second graph where the speed of the clock is 0.866 and gamma equals two. The formula says that the time on the clock will be one half of the coordinate time and we see that because the spacing of the dots is twice the spacing of the grid lines for the graph. In other words, the first dot after the origin, representing a clock time of one second occurs at a coordinate time of two seconds.

Got it?
Attached Thumbnails
One object stationary.png   One object moving.png  
John Huang
John Huang is offline
#82
Dec12-12, 08:12 AM
P: 52
Quote Quote by tensor33 View Post
For the last time, LT is a part of SR.
If I ignore the fallowing logic, then, yes, LT is part of SR.

Quote Quote by tensor33 View Post
In special relativity (or, hypothetically far from all gravitational mass), clocks that are moving with respect to an inertial system of observation are measured to be running more slowly. This effect is described precisely by the Lorentz transformation.
This is the only way SR can include LT, by CLAIMING.


Quote Quote by tensor33 View Post
And you were right about Δt' = γ(Δt-(vΔx/c^2)) but wrong about Δt' = Δt/γ. When Δx=0, Δt'=γΔt not Δt/γ. I don't understand where you got Δt' = Δt/γ. I've never seen this equation used and it seems to be what is getting in your way.
Think about it, HOW the observers in the stationary system S measure the "event period" of ONE SPECIFIC event? By the stationary clock S-CLOCK in S, right? For the observers in the moving system S', they will use the moving clock M-CLOCK to measure the "event period" of ONE SPECIFIC event. For the SPECIFIC event in the arranged situation the observers in S will measure only one "event period" Δt and for observers in S' the observers will measure only one "event period" Δt'. Am I correct?

In the SPECIFIC situation, SR states based on the relative speed "v", the relation of the speed of time should be Δt' = Δt/γ, but regarding the "event period", we should go by Δt' = γ(Δt-(vΔx/c^2)). The "event period" reflects the speed of time, isn't it?
Doc Al
Doc Al is offline
#83
Dec12-12, 08:18 AM
Mentor
Doc Al's Avatar
P: 40,905
Quote Quote by John Huang View Post
If I ignore the fallowing logic, then, yes, LT is part of SR.

This is the only way SR can include LT, by CLAIMING.
Enough already. Please define what you mean by "SR".

Everyone in this thread besides you understands that SR means Special Relativity. And of course Special Relativity includes the Lorentz transformations. Any claim otherwise is just silly.

Please give your definition of "SR" in your very next post. (I am giving you the benefit of the doubt here.)
Doc Al
Doc Al is offline
#84
Dec12-12, 08:30 AM
Mentor
Doc Al's Avatar
P: 40,905
Quote Quote by John Huang View Post
Think about it, HOW the observers in the stationary system S measure the "event period" of ONE SPECIFIC event? By the stationary clock S-CLOCK in S, right? For the observers in the moving system S', they will use the moving clock M-CLOCK to measure the "event period" of ONE SPECIFIC event. For the SPECIFIC event in the arranged situation the observers in S will measure only one "event period" Δt and for observers in S' the observers will measure only one "event period" Δt'. Am I correct?
Events happen at a single instant. There is no Δt for a single event, only between two events. "Event period" makes no sense.
In the SPECIFIC situation, SR states based on the relative speed "v", the relation of the speed of time should be Δt' = Δt/γ, but regarding the "event period", we should go by Δt' = γ(Δt-(vΔx/c^2)).
Once again, the Lorentz transformations apply to any time interval. So Δt' = γ(Δt-(vΔx/c^2)) is always true. Realize that that is only one of the four basic transformations; Another is Δt = γ(Δt' + (vΔx'/c^2))

The time dilation formula Δt' = Δt/γ is only true for the special case where the events happen at the same position in the primed frame and thus Δx' = 0. In words, the time dilation formula says that 'moving clocks run slow'.
The "event period" reflects the speed of time, isn't it?
No.
jtbell
jtbell is offline
#85
Dec12-12, 08:56 AM
Mentor
jtbell's Avatar
P: 11,255
Quote Quote by John Huang View Post
HOW the observers in the stationary system S measure the "event period" of ONE SPECIFIC event?
Quote Quote by Doc Al View Post
"Event period" makes no sense.
If you [John] don't understand the standard terminology that we use when we discuss SR, it doesn't help to make up your own terminology that we don't understand.
DaleSpam
DaleSpam is offline
#86
Dec12-12, 10:17 AM
Mentor
P: 16,485
Quote Quote by John Huang View Post
I could make a new term, like a "story", to include all continuous events relative to an object. If the object is stationary in a system, then, that object has a "stationay story" and if the object moves in the system then that object has a "moving story".

However, If LT is part of SR, then, this will not be a problem any more.

SR can let the LT portion of SR handle all "moving stories". Then the famous time dilation equation of SR can handle all "stationary stories" in the moving system. But for a "stationary story" in the stationary system people cannot use the famous time dilation equation of SR to calculate the "event period". They must use the inverse equation of it.

Or, people can simply use LT to calculate the event periods (or story periods) for all kinds of story in both systems.
This is essentially correct. The standard term for "story" is "worldline".

For pedagogical reasons my recommendation is always to follow your last suggestion which I highlighted in bold. It automatically simplifies to the time dilation equation whenever it is appropriate and it avoids accidentally using it when it is not appropriate.
DaleSpam
DaleSpam is offline
#87
Dec12-12, 10:20 AM
Mentor
P: 16,485
Quote Quote by John Huang View Post
I think, after that, SR cannot claim to include LT any more.
SR still includes the LT. A specialized simplification of a general equation does not in any way invalidate the general equation.
John Huang
John Huang is offline
#88
Dec12-12, 11:39 PM
P: 52
Quote Quote by Doc Al View Post
Events happen at a single instant. There is no Δt for a single event, only between two events. "Event period" makes no sense.
Thanks for your correction. Now, I will rewrite my sentence.

How do observers everywhere in the stationary system S measure the "time period" Δt of ONE SPECIFIC SECTION of continuous events from event 1 at point A to event 2 at point B? They use the synchronized stationary clocks in S. For the observers everywhere in the moving system S', they will use the synchronized moving clocks to measure the "time period" Δt' from event 1 to event 2. Now, we have a better defined case.

There are two ways to do the measure jobs.

The easy way is to measure event time t1 and t1' for event 1 at point A and measure event time t2 and t2' for event 2 at point B. We will have Δt' = t2'-t1' and Δt = t2-t1.

The difficult way is to measure event time by assigned observers, one in S and one in S'. The simplest assginment is let them stay at origin points O and O'. Let us look at this simplest situation. Since the distances of each pair of four points A, B, O and O' could be different, we must adjust the influence of the distance. When A=B=O', it will be the situation arranged by Einstein. We will have Δt' = t2'-t1' and Δt = t2-t1.

For this SPECIFIC SET of t1,t2, t1' and t2', SR states based on the relative speed "v", the relation of the speed of time in S' and S should be Δt' = Δt/γ, but regarding the "time period", we should go by Δt' = γ(Δt-(vΔx/c^2)). Am I correct?

Quote Quote by Doc Al View Post
Once again, the Lorentz transformations apply to any time interval. So Δt' = γ(Δt-(vΔx/c^2)) is always true. Realize that that is only one of the four basic transformations; Another is Δt = γ(Δt' + (vΔx'/c^2))

The time dilation formula Δt' = Δt/γ is only true for the special case where the events happen at the same position in the primed frame and thus Δx' = 0. In words, the time dilation formula says that 'moving clocks run slow'.
Yes, "In words, the time dilation formula says that 'moving clocks run slow'." and for relative speed v, the time dilation formula Δt' = Δt/γ stands.

In Δt' = Δt/γ, Δt' is for the time period in the moving system S' of a given SECTION OF EVENTS and Δt is for the stationary system S of the same SECTION OF EVENTS.
John Huang
John Huang is offline
#89
Dec12-12, 11:41 PM
P: 52
Quote Quote by ghwellsjr View Post
Got it?
Thanks.
harrylin
harrylin is offline
#90
Dec13-12, 05:02 AM
P: 3,178
Quote Quote by John Huang View Post
Thanks for your correction. Now, I will rewrite my sentence.

How do observers everywhere in the stationary system S measure the "time period" Δt of ONE SPECIFIC SECTION of continuous events from event 1 at point A to event 2 at point B? They use the synchronized stationary clocks in S. For the observers everywhere in the moving system S', they will use the synchronized moving clocks to measure the "time period" Δt' from event 1 to event 2. Now, we have a better defined case.
That's much better defined.
There are two ways to do the measure jobs.

The easy way is to measure event time t1 and t1' for event 1 at point A and measure event time t2 and t2' for event 2 at point B. We will have Δt' = t2'-t1' and Δt = t2-t1.
The difficult way is to measure event time by assigned observers, one in S and one in S'. The simplest assginment is let them stay at origin points O and O'.
OK so you now moved on to two ways of measuring the same interval: the simplest, "local" way of measuring and the indirect or "non-local" way. SR predicts the same for both set-ups; that is necessary for consistency.
Let us look at this simplest situation. Since the distances of each pair of four points A, B, O and O' could be different, we must adjust the influence of the distance. When A=B=O', it will be the situation arranged by Einstein.
Ok, you take here the special case that Δx'=0, just as you did before - correct? And for the LT we choose O=O' at t1=0. Here's a generic sketch of it:

t1................t2
O--------------------- S
A.................B

A,B
O'---------------- S'
t1'
t2'

We will have Δt' = t2'-t1' and Δt = t2-t1.

For this SPECIFIC SET of t1,t2, t1' and t2', SR states based on the relative speed "v", the relation of the speed of time in S' and S should be Δt' = Δt/γ, but regarding the "time period", we should go by Δt' = γ(Δt-(vΔx/c^2)). Am I correct? [..]
That is not correct; Δt' = γ(Δt-(vΔx/c^2)) is valid for both; and there is no difference of prediction between using local clocks or distant observers. I copy-paste my earlier reply:

1. For x1'=x2' (Δx'=0, clock at rest in S', moving in S): Δt' = Δt/γ
2. For x1=x2 (Δx=0, clock at rest in S, moving in S'): Δt' = γΔt

Once more: you selected the time period of situation 1, with x1'=x2' and thus Δx≠0.
You should find from the LT that the time period Δt' = Δt/γ for that situation, based on the relative speed v.
If that is not clear to you, please ask.


Register to reply

Related Discussions
Einstein's Theory of Special relativity Special & General Relativity 15
How would you explain the Special Theory of Relativity using Physics? Special & General Relativity 10
Can someone explain Einstein's relativity theory? Special & General Relativity 69
Einstein's theory of Special Relativity Special & General Relativity 8