Lorentz contraction from space-time interval invariance

[bernhard.rothenstein]

Please tell me if it is possible to derive the formula which accounts for the Lorentz contraction from the invariance of the space-time interval.
Thanks

 

[neutrino]

I don't know, maybe I'm blabbering (it's pretty late here), but I don't see how we can relate "space-components" (of two different inertial frames) of the spacetime interval this way. For one, the interval relates the space and time coordinates of two distinct events as observed in two different frames. While measuring the length say, of a rod, \Delta x must be measured at one instant in time, and therefore \Delta t must be 0. Since these events are simultaneous in the unprimed frame, they must not be so in the primed frame (Non-zero \Delta t'), which implies \Delta x' was not measured at a "single point" in time.

Someone please tell if that made any sense, and please correct me if I'm wrong. Something tells me that there might be a way to derive the formula, since the invariance is a fundamental property, but as I said, I'm getting ready for some Z's now. :zzz:

 

[morrobay]

Is this the algebra your looking for ?

Let " equal squared
Let 0" -(cT)" equal spacetime interval of stationary observer
Let (vt)" - (ct)" equal spacetime interval of moving observer
-(cT)" = (vt)" -(ct)" invariance of spacetime interval
c"t" = v"t" + c"T" divide by t"
c"=v"+ c"T"/t" divide by c"
1-v"/c" =T"/t" take sq root both sides
(sq rt) 1-v"/c" = T/t
t= T/ (sq rt) 1-v"/c"

 

[bernhard.rothenstein]

length contraction from space-time interval invariance

Is this the algebra your looking for ?

Let " equal squared
Let 0" -(cT)" equal spacetime interval of stationary observer
Let (vt)" - (ct)" equal spacetime interval of moving observer
-(cT)" = (vt)" -(ct)" invariance of spacetime interval
c"t" = v"t" + c"T" divide by t"
c"=v"+ c"T"/t" divide by c"
1-v"/c" =T"/t" take sq root both sides
(sq rt) 1-v"/c" = T/t
t= T/ (sq rt) 1-v"/c"

Thanks for your answer. That is the derivation of the time dilation formula.
My problem is how to find out
L=L(0)sqrt(1-VV/cc)
from the invariance of the space time interval without to use the previous result.
Regards

 

[yuiop]

Try this:

-(c^2T^2) +X^2 = -c^2t^2 + L_0^2

Say a clock is transported from A to B (distance L_0) in time t at velocity v relative to your frame.
T is the proper time of the transported clock as measured by itself (by definition).
X is the proper distance the clock moves in its own frame which is zero .

-(c^2T^2) = -c^2t^2 + L_0^2

T^2 = t^2 - L_0^2/c^2

T^2 = L_0^2/v^2 - L_0^2/c^2

T^2 v^2 = L_0^2 -L_0^2v^2/c^2

T^2 v^2 = L_0^2 (1-v^2/c^2)

vT = L_0\sqrt{1-v^2/c^2} = L

L is the distance A to B as measured by an observer comoving with the clock.

 

[country boy]

Begin with the unit invariant interval x^2-(ct)^2=1. This traces out a hyperbola on a Minkowski diagram that gives the unit distance along the x-axis and any other x'-axis. Now place one end of a rod of length L at the origin and the other end on the x-axis. This rod is at rest on the x-axis. The world lines of the ends of L cross the x'-axis at the origin and some other point L'. To get the Lorentz contraction, compare L and L' to where the hyperbola crosses the x- and x'-axes (the unit lengths in the two frames).

 

[bernhard.rothenstein]

space-time interval invariance

Try this:

-(c^2T^2) +X^2 = -c^2t^2 + L_0^2

Say a clock is transported from A to B (distance L_0) in time t at velocity v relative to your frame.
T is the proper time of the transported clock as measured by itself (by definition).
X is the proper distance the clock moves in its own frame which is zero .

-(c^2T^2) = -c^2t^2 + L_0^2

T^2 = t^2 - L_0^2/c^2

T^2 = L_0^2/v^2 - L_0^2/c^2

T^2 v^2 = L_0^2 -L_0^2v^2/c^2

T^2 v^2 = L_0^2 (1-v^2/c^2)

vT = L_0\sqrt{1-v^2/c^2} = L

L is the distance A to B as measured by an observer comoving with the clock.

Thanks. Do you know a derivation of the invariance of the space-time interval

 

[1effect]

Please tell me if it is possible to derive the formula which accounts for the Lorentz contraction from the invariance of the space-time interval.
Thanks

Look up pages 42-43 in "Spacetime Physics" by Wheeler and Taylor.

 

[1effect]

Thanks. Do you know a derivation of the invariance of the space-time interval

That is a postulate, so there is no derivation for it.

 

[yuiop]

Thanks. Do you know a derivation of the invariance of the space-time interval

I notice in another thread https://www.physicsforums.com/archive/index.php/t-115451 you posted this:

Imagine systems S and S'. Person A stands at the origin in S and person B stands in the origin of S', which moves with velocity v relative to S. The origins coincide at t=t'=0. Imagine A lights a spark at t=0. A spherical light wave spreads out in all directions. At any point in time A stands in the center of the sphere which has radius ct. So the wavefront obeys the equation:
x^2+y^2+z^2=(ct)^2
Same thing holds in S'. B also stands in the center of the sphere at all times (2nd postulate):
x'^2+y'^2+z'^2=(ct')^2
so whatever happens to the coordinates, it should always be the case that:
x^2+y^2+z^2-(ct)^2=x'^2+y'^2+z'^2-(ct')^2

..which seems perfectly valid to me, but you were bothered that the final equation is obtained by subtracting (ct)^2 from both sides of the first two equations so that the first two equations equal zero and then saying if two quantities are equal to zero then they are equal to each other.

As far as know using the technique of setting two equations to zero and saying they equal to each other is a perfectly normal mathematical technique.

 

[bernhard.rothenstein]

space time interval invariance

I notice in another thread https://www.physicsforums.com/archive/index.php/t-115451 you posted this:



..which seems perfectly valid to me, but you were bothered that the final equation is obtained by subtracting (ct)^2 from both sides of the first two equations so that the first two equations equal zero and then saying if two quantities are equal to zero then they are equal to each other.

As far as know using the technique of setting two equations to zero and saying they equal to each other is a perfectly normal mathematical technique.

Thanks. Do you consider that starting with
c*t*-x*=d* (1) *2
c"t'*-x'*=d* (2)
a consequence of the invariance of distances measured perpendicular to the direction of relative motion d then
c*t*-x*=c*t'*-x'*
is more convincing?
Regards

 

[country boy]

By setting two zero-interval expressions equal to each other you are not demonstrating the in variance for non-zero values. The argument could be completed by deriving the Lorentz transformations from your two light spheres and then showing that all spacetime intervals are invariant.

 

[yuiop]

Thanks. Do you consider that starting with
c*t*-x*=d* (1) *2
c"t'*-x'*=d* (2)
a consequence of the invariance of distances measured perpendicular to the direction of relative motion d then
c*t*-x*=c*t'*-x'*
is more convincing?
Regards

Hi Bernhard. Your symbols are little confusing and I'm not sure exactly what you are getting at. What is the stray *2 at the end of equation (1) ? I'm assuming the bracketed number are equation numbers and not used in the same context as you you L(0) earlier. In equation (2) I'm assuming you meant to put c* instead of c" to represent c^2. I'm not sure where you get d^2 and I'm not sure that starting with the premis that distances measured perpendicular to the direction of relative motion is invarient demonstrates that the interval is invarient in all directions is ... entirely convincing.

Since ct = x and c^2t^2 = x^2 the quantity c^2t^2-x^2 is zero and equating it to d^2 only hides the zero.

The statement by countryboy

By setting two zero-interval expressions equal to each other you are not demonstrating the in variance for non-zero values. The argument could be completed by deriving the Lorentz transformations from your two light spheres and then showing that all spacetime intervals are invariant.

would appear to be correct.

 

[yuiop]

This is my stab at deriving the invarient interval but it's not very rigorous ;)


Start with speed of light is the same to all observers, so

(1) x/t = c and
(2) x'/t' =c

(1) and (2) are in terms of velocity so re-arrange them in terms of distance (because we are looking for the invarient spatial interval) to get

(3) x = ct and
(4) x' = ct'

Now x can take positive or negative values depending on direction from the origin so square both sides of each equation to remove any pesky negatives. (If we were doing the full version with x, y and z coordinates there would be a sqrt(x^2+y^2+z^2) term that can have a positive or negative root)

(5) x^2 = (ct)^2 and
(6) x'^2 = (ct')^2

Now subtract the left side of (6) from the left side of (5) and right side of (6) from the right side of (5). We are in effect subtracting the same value from each side of equation (5) because the equality of (6)

(7) x^2-x'^2 = (ct)^2 - (ct')^2

Rearrange to get

(8) x^2 - (ct)^2 = x'^2 -(ct')^2 which is what we are looking for.

The full version can be done by starting with

(9) sqrt(x^2+y^2)+z^2)/t = c and
(10) sqrt(x'^2+y'^2+x'^2)/t' =c

but I'm too lazy :P

 

[bernhard.rothenstein]

space-time interval invariance

Thanks kev. It is of big help to me.
Please have a critical look at the following:
Consider the well known light clock experiment performed in its rest frame I(0). Let d be the distance between the two mirrors. Being measured along a direction which is perpendicular to the direction of relative motion between the involved inertial reference frames it has the same magnitude in all of them. Consider the same experiment from I relative which moves relative to I(0) in the standard way. Let x be the horizontal displacement of the upper mirror and ct the distance traveled by the light signal between the two mirrors. Pythagoras' theorem requires
d^2=c^2t^2-x^2 (1)
Invariance of d requires the invariance of the right side of (1) and so in a I' reference frame we should have
d^2=c^2t'2-x'^2. (2)
Equating two different from zero relativistic expressions we have
c^2t^2-x^2=c^2t'^2-x'^2 (3)
Do you consider that (3) is a prove for the invariance of the relativistic space-time interval?

 

[yuiop]

Thanks kev. It is of big help to me.
Please have a critical look at the following:
Consider the well known light clock experiment performed in its rest frame I(0). Let d be the distance between the two mirrors. Being measured along a direction which is perpendicular to the direction of relative motion between the involved inertial reference frames it has the same magnitude in all of them. Consider the same experiment from I relative which moves relative to I(0) in the standard way. Let x be the horizontal displacement of the upper mirror and ct the distance traveled by the light signal between the two mirrors. Pythagoras' theorem requires
d^2=c^2t^2-x^2 (1)
Invariance of d requires the invariance of the right side of (1) and so in a I' reference frame we should have
d^2=c^2t'2-x'^2. (2)
Equating two different from zero relativistic expressions we have
c^2t^2-x^2=c^2t'^2-x'^2 (3)
Do you consider that (3) is a prove for the invariance of the relativistic space-time interval?

OK, I see what you are getting at now.
When the light clock is moving relative to you the light pulse goes a distance ct in time t and this is equivalent to the diagonal distance sqrt(d^2+x^2) where the light clock moves a distance x in the same time t. This works for any relative velocity of the light clock so its seems a reasonable approach. However I'm not qualified to say if this would be regarded as a rigorous formal proof ;)

 

[1effect]

Do you consider that (3) is a prove for the invariance of the relativistic space-time interval?

The invariance of the space-time interval is a postulate. You don't prove postulates since they are ...postulates.

 

[yuiop]

The invariance of the space-time interval is a postulate. You don't prove postulates since they are ...postulates.

Is it a postulate? Minkowski came up with the invariant interval and his spacetime after Einstein formulated Special Relativity which has the invariance of the speed of light and invariance of the laws of physics in any inertial reference frame. Eistien did not build special relativity on the postulate of the invariant interval. (Well as far as I know)

 

[bernhard.rothenstein]

invariance of space-time interval

The invariance of the space-time interval is a postulate. You don't prove postulates since they are ...postulates.

When I started to learn English helped by the British Broadcasting Company (BBC) professor Grammar told me that "English is a very flexible language". I think that special relativity is a very flexible theory and you can derive almost every thing starting with something that is in accordance with the two postulates or is a consequence of it. I have seen papers in which the Lorentz transformations are derived from time dilation, from length contraction, from both of them, from the addition law of parallel speeds, from the Doppler Effect, from the invariance of the space-time interval. I have not the competence to decide which is the best way to follow. I think that the invariance of the Minkowski distance is a good way for doing it, but only after having proved its invariance and that is in my opinion the weak point of Taylor's approach to SR.
Thanks for your contribution to my thread.

 

[1effect]

Is it a postulate? Minkowski came up with the invariant interval and his spacetime after Einstein formulated Special Relativity which has the invariance of the speed of light and invariance of the laws of physics in any inertial reference frame. Eistien did not build special relativity on the postulate of the invariant interval. (Well as far as I know)

Minkowski did. It is a postulate in Mikowski's formalism, it is a theorem in Einstein's 1905 paper. But then, Einstein derives it as a consequence of the Lorentz transforms. The OP asked for the opposite order in deriving the Lorentz transforms from the invariance of the interval , so , something has to be the starting postulate.

 

[bernhard.rothenstein]

is there a link between the invariance of the interval and the clock synchronization in the involved inertial reference frames?

 

[snoopies622]

I thought that the invariance of the speed of light is supposed to be the only postulate.

 

[dx]

Let an observer (2) be positioned on one side of a rod of length l_0 in his rest frame. Let (1) be another observer relative to whom (2) is moving to the left with velocity v. Their initial configuration is shown below in (1)'s frame.

(3)...l...(2)
(1)

the rod is moving leftwards.

Let the length of the rod be l in (1)'s frame.

E_1 : (3) and (1) coincide.
E_2 : (2) and (1) coincide.

According to (1), the time between the two events is \frac{l}{v}, and the distance is zero. Therefore,

t_1^2 - x_1^2 = \frac{l^2}{v^2}

According to (2), the time between the two events is \frac{l_0}{v}, and the distance is l_0. Therefore,

t_2^2 - x_2 ^2 = \frac{l_0^2}{v^2} - l_0^2

Equating the two, we get

l = l_0\sqrt{1-v^2}

 

[Dale]

Nice!

 

[bernhard.rothenstein]

lengths contraction and clock synchronization

Let an observer (2) be positioned on one side of a rod of length l_0 in his rest frame. Let (1) be another observer relative to whom (2) is moving to the left with velocity v. Their initial configuration is shown below in (1)'s frame.

(3)...l...(2)
(1)

the rod is moving leftwards.

Let the length of the rod be l in (1)'s frame.

E_1 : (3) and (1) coincide.
E_2 : (2) and (1) coincide.

According to (1), the time between the two events is \frac{l}{v}, and the distance is zero. Therefore,

t_1^2 - x_1^2 = \frac{l^2}{v^2}

According to (2), the time between the two events is \frac{l_0}{v}, and the distance is l_0. Therefore,

t_2^2 - x_2 ^2 = \frac{l_0^2}{v^2} - l_0^2

Equating the two, we get

l = l_0\sqrt{1-v^2}

thanks for the derivation. please tell me if it involves clock synchronization the times involved being displayed by Einstein synchronized clocks.

 

[Dale]

No synchronization involved. The proper length can be measured with the two-way speed of light using a single clock. The other frame doesn't measure any distance, only a time at a single location, so again no synchronization.

 

[Mentz114]

Dx,
nice derivation. I take it all times are measured in 1's frame.

M

[edit] this post coincided with DaleSpam's ( according to my clock !).

 

[dx]

Dx,
nice derivation. I take it all times are measured in 1's frame.

t_1 is measured in 1's frame. t_2 is measured in 2's frame.

 

[dx]

thanks for the derivation. please tell me if it involves clock synchronization the times involved being displayed by Einstein synchronized clocks.

I'm not sure what you mean. The expression for the interval which I used was \tau^2 = t^{2} - x^{2}. We must use Einstein's simultaneity convention when we measure t and x. If we use some other convention, then the expression for the interval will be different.

 

[bernhard.rothenstein]

interval derivation

That is a postulate, so there is no derivation for it.

o.k. the interval contains the times t and t' and works even for the time intervals dt and dt'. It would not be necessary to mention the clocks that display those times, how are they synchronized...? I think that before to postulate Einstein's clock synchronization should be mentioned.

 

[DrGreg]

Thanks. Do you know a derivation of the invariance of the space-time interval

In case you never saw it (or forgot), take a look at my post here (due to Bondi).