Is relativistic effect of length contraction physically real ?

[kahoomann]

Is relativistic effect of length contraction physically "real"?

Is Lorentz contraction a real contraction? For example, if one tries to accelerate a solid body, does its contraction require an extra input of energy to squeeze the atoms of the body closer together? Will this extra energy go into the total mass of the moving body?

 

[Fredrik]

It's very real, but not in that sense. This should be obvious if you consider that it doesn't matter if it's the object or you who changed velocity.

 

[Dale]

It is physically real, this can be seen by the fact that particle accelerators require relativistic corrections to the "bunch length" in order to determine the interactions of the particles.

Lorentz contraction is strain-free, as can be measured by a strain gauge, so it does not require additional energy. Don't forget that the fields around an atom also length contract.

 

[granpa]

Is Lorentz contraction a real contraction?

yes and no.

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first.

https://www.physicsforums.com/showthread.php?t=236978

 

[kahoomann]

yes and no.

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first.

https://www.physicsforums.com/showthread.php?t=236978

The key question is:
if one tries to accelerate a solid body, does its contraction require an extra input of energy to squeeze the atoms of the body closer together?

 

[russ_watters]

That question, already answered, is no. An object does not contract in its own frame.

 

[kahoomann]

Bernard Schutz, in his book, indicates that Lorentz contraction does require an extra input of energy to squeeze the atoms of the body closer together. Is there anyone who agrees with him?

see this link
http://books.google.com/books?id=jR...a+solid+body"&sig=T8L2gCi4h6HUs1QKBq60HBbU-yM

 

[Fredrik]

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first.

This is completely false. A describes B the same way B describes A.

The key question is:
if one tries to accelerate a solid body, does its contraction require an extra input of energy to squeeze the atoms of the body closer together?

That question was answered twice before you asked it again, but I'll say it more clearly: If the acceleration is linear, then the answer is definitely no.

If you accelerate a real object by pushing it at one end, you will compress it a bit, but if you don't break it, every microscopic piece of it will restore itself to its original rest length in co-moving inertial frames. This will heat the object a bit, so the work you perform when you push the rear of the object doesn't get converted to forward motion with 100% efficiency. This is an effect that doesn't really have anything to do with relativity, so I assume that this isn't what you had in mind.

A Lorentz contraction is real in the sense that objects really do get shorter or longer when your velocity relative to the object changes (regardless of whether it was you or the object that accelerated). It's not just that that they appear to get shorter or longer. The reason why lengths change is that your velocity is what determines which 3-dimensional "slice" of space-time you will consider space. (There's nothing more important than this in all of SR, so you should try really hard to understand it if you're at all interested). Two observers who measure the length of the same object will disagree because they are measuring the lengths of different paths in space-time.

So why did I say "if the acceleration is linear..."? Because there are situations where it just isn't possible for each microscopic piece to restore itself to its original length in co-moving inertial frames. The simplest example is a rotating disc. When you give a wheel a spin, the material will be forcefully stretched everywhere along the circumference by a factor that exactly compensates for the Lorentz contraction. So in this case we are performing additional work, not to cause the Lorentz contraction but to make sure that lengths remain the same when they do get Lorentz contracted.

 

[granpa]

Originally Posted by granpa View Post

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first.

This is completely false. A describes B the same way B describes A.

how does that contradict what i said??

 

[Fredrik]

Bernard Schutz, in his book, indicates that Lorentz contraction does require an extra input of energy to squeeze the atoms of the body closer together

see this link
http://books.google.com/books?id=jR...a+solid+body"&sig=T8L2gCi4h6HUs1QKBq60HBbU-yM

I really like Schutz because I think the SR section in his "A first course on general relativity" is awesome, but this is just wrong. One way to see that is to consider the acceleration of a single classical point particle. It's energy will increase by $\gamma mc^2-mc^2$, and it's definitely not because atoms are being squeezed together. See e.g. the recent thread about derivations of E=mc².

 

[Dale]

a moving object and a stationary object can't both be shorter than the other.

Sure they can, draw a spacetime diagram and see!

 

[Fredrik]

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first.

This is completely false. A describes B the same way B describes A.

how does that contradict what i said??

I suppose I could have expressed myself more clearly. I think you could have too, because I'm not sure what you were saying there. You appear to be saying that one object shrinks and one doesn't. That's not what happens. If A and B are two identical objects moving at the same velocity and you change the velocity of one of them, the following will be the result regardless of whether it was A or B that accelerated: A is shorter in B's frame. B is shorter in A's frame. Nothing has really changed about either of the objects. They just disagree about which slices of space-time are space.

 

[granpa]

a moving object and a stationary object can't both be shorter than the other. 1 really shrinks. the other only APPEARS to shrink when viewed by the first due to loss of simultaneity. you obviously didnt even look at the link i posted.

 

[Fredrik]

you obviously didnt even look at the link i posted.

You're right about that. You said something that was clearly incorrect and posted a link to what was obviously another thread without explaining the reason. It didn't seem worth the effort to click it.

a moving object and a stationary object can't both be shorter than the other.

Not in one frame, that's true, but if that's what you mean, you should say it.

1 really shrinks. the other only APPEARS to shrink when viewed by the first due to loss of simultaneity.

But the first also "appears to shrink" when viewed by the other, for the same reason. So if "1 really shrinks", then so does the other.

 

[granpa]

But the first also "appears to shrink" when viewed by the other, for the same reason.

thats where you are wrong. its not for the same reason, as i clearly showed and proved mathematically in the link i posted.

but it is impossible to tell which is real and which isnt.

 

[MeJennifer]

Is Lorentz contraction a real contraction? For example, if one tries to accelerate a solid body, does its contraction require an extra input of energy to squeeze the atoms of the body closer together? Will this extra energy go into the total mass of the moving body?

I think it a good idea to differentiate between the observation of the length of a rod in motion with respect to an observer and the effects of acceleration on a pulled or pushed rod. Those are two different things.

 

[Dale]

but it is impossible to tell which is real and which isnt.

It is easy to tell which is real: they both are.

 

[granpa]

It is easy to tell which is real: they both are.

you believe that a can be shorter than b while b is at the same time shorter than a? ok. then perhaps you can explain to me where my calculations went wrong in this thread:
https://www.physicsforums.com/showthread.php?t=236978

 

[matheinste]

Hello granpa

Quote:-

----but it is impossible to tell which is real and which isnt.----

I would interpret this as saying that you cannot tell which contraction is real and which is illusory. So we cannot tell which is which. This is a bit like the definition of equality. This means they are the same. That means they are both real or both illusory.

Matheinste.

 

[granpa]

i should have said that it is impossible to tell which is due to the length contraction of the object being viewed and which is due to loss of simultaneity in the observer.

 

[matheinste]

Quote:-

---- loss of simultaneity in the observer.---

What does that mean

Mateinste

 

[Dale]

you believe that a can be shorter than b while b is at the same time shorter than a? ok. then perhaps you can explain to me where my calculations went wrong in this thread:
https://www.physicsforums.com/showthread.php?t=236978

Sure, you didn't use the full Lorentz transform, just bits and pieces cobbled together.

Look at the attached diagram which is simply the Lorentz transform for an unprimed "rest" frame (black) and a primed frame (white) moving with relative v=0.6c (c=1). If the back end of a proper-length 2 rod at rest in the unprimed frame follows the worldline x=0 then the front end follows the worldline x=2. If the back end of a proper-length 2 rod at rest in the primed frame follows the worldline x'=0 then the front end follows the worldline x'=2.

Note that, starting from the origin, the t=0 line intersects the x'=2 line before the x=2 line. The primed rod is shorter than the unprimed rod (in the primed frame).

Note that, starting from the origin, the t'=0 line intersects the x=2 line before the x'=2 line. The unprimed rod is shorter than the primed rod (in the unprimed frame).

There is no contradiction, they are talking about two different sets of events.

 

[Doc Al]

you believe that a can be shorter than b while b is at the same time shorter than a?

When you say "a is shorter than b" you must specify the frame in which measurements are being made. Until you do, the statement is meaningless. According to frame b, frame a lengths are "contracted"; according to frame a, frame b lengths are. No problem.

 

[granpa]

Sure, you didn't use the full Lorentz transform, just bits and pieces cobbled together.

Look at the attached diagram which is simply the Lorentz transform for an unprimed "rest" frame (black) and a primed frame (white) moving with relative v=0.6c (c=1). If the back end of a proper-length 2 rod at rest in the unprimed frame follows the worldline x=0 then the front end follows the worldline x=2. If the back end of a proper-length 2 rod at rest in the primed frame follows the worldline x'=0 then the front end follows the worldline x'=2.

Note that, starting from the origin, the t=0 line intersects the x'=2 line before the x=2 line. The primed rod is shorter than the unprimed rod (in the primed frame).

Note that, starting from the origin, the t'=0 line intersects the x=2 line before the x'=2 line. The unprimed rod is shorter than the primed rod (in the unprimed frame).

There is no contradiction, they are talking about two different sets of events.

thats obviously true but what does that have to do with what i said? it contradicts nothing i have said.

 

[granpa]

Quote:-

---- loss of simultaneity in the observer.---

What does that mean

Mateinste

if you are asking how loss of simultaneity changes ones measurement of other objects then you should read this:
https://www.physicsforums.com/showthread.php?t=236978

if you are asking what i mean by 'in the observer' then i mean that the atoms that make up the observer are interacting by means of electric and magnetic fields that themselves change at the speed of light and this leads to a shift of simultaneity (as that observer would define it).

 

[Fredrik]

If you multiply every vector in DaleSpam's diagram with

$$\gamma\begin{pmatrix}1 && -v\\-v && 1\end{pmatrix}$$

The diagram will turn into its mirror image with black and white swapped, i.e. the white lines will look the way the black lines do now, and the black lines will look like the white lines do now except they'll be tilted the other way.

It makes no sense to say that one of these diagrams represents reality and that the other doesn't.

 

[Dale]

thats obviously true but what does that have to do with what i said?

It directly addresses your concerns about both being real (a is shorter than b and b is shorter than a).

 

[Dale]

If you multiply every vector in DaleSpam's diagram with

$$\gamma\begin{pmatrix}1 && -v\\-v && 1\end{pmatrix}$$

The diagram will turn into its mirror image with black and white swapped, i.e. the white lines will look the way the black lines do now, and the black lines will look like the white lines do now except they'll be tilted the other way.

It makes no sense to say that one of these diagrams represents reality and that the other doesn't.

This is correct. Because of that, some people draw both frames at the same angle (i.e. from the perspective of a "midpoint" observer) to emphasize the symmetry. This was a drawing that I did rather early on in my understanding of SR, so I wasn't skilled enough at the time to do such a symmetric diagram. But even without making the symmetry obvious you can still see how each frame sees the other as length contracted and time dilated.

 

[matheinste]

Hello granpa

Quote:-

---if you are asking what i mean by 'in the observer' then i mean that the atoms that make up the observer are interacting by means of electric and magnetic fields that themselves change at the speed of light and this leads to a shift of simultaneity (as that observer would define it).----

I'm afraid i'll have to drop out at this point. The above is out of my depth.

Matheinste

 

[granpa]

It directly addresses your concerns about both being real (a is shorter than b and b is shorter than a).

its just a mathematical diagram of exactly what i just said. it shows that one objects measurement of the length of the other depends on that objects length and on the definition of simultaneity. it in no way at all contradicts anything i said nor anything in this:
https://www.physicsforums.com/showthread.php?t=236978

 

[granpa]

If you multiply every vector in DaleSpam's diagram with

$$\gamma\begin{pmatrix}1 && -v\\-v && 1\end{pmatrix}$$

The diagram will turn into its mirror image with black and white swapped, i.e. the white lines will look the way the black lines do now, and the black lines will look like the white lines do now except they'll be tilted the other way.

It makes no sense to say that one of these diagrams represents reality and that the other doesn't.

well your math is out of my league but let me say this. it is perfectly true, and i have said it many times, that every observer considers himself to be stationary and every observer sees exactly the same thing (that he would see if he were indeed stationary). this is perfectly symmetrical. but the equations of relativity show that they see the same thing for entirely different reasons. some see an object to be shortened because that object is length contracted and others see objects to be shortened because the observer is experiencing a loss of simultaneity. this is not symmetrical at all and fancy math manipulation doesn't change that.

 

[Fredrik]

If you think so, it must be because you're focusing on the wrong parts of the math. For example, I don't think you have fully understood the very intimate relationship between relativity of simultaneity and length contraction. Relativity of simultaneity is precisely what makes different observers disagree about what events the endpoints are when they measure the length of the same object, and that's the reason they measure different lengths. So you can actually think of the simultaneity stuff as the cause of length contraction. But it's probably better to think of the Minkowski metric, or equivalently, the Lorentz transformation, as the cause of all of these things.

 

[granpa]

loss of simultaneity can explain some length contraction but not all. that's the point.Relativity of simultaneity is precisely what makes different observers disagree about what events the endpoints are when they measure the length of the same object,

thats exactly what i explained here:
https://www.physicsforums.com/showthread.php?t=236978
has nobody ever bothered to read it?

 

[Mentz114]

granpa,
Relativity of simultaneity is a consequence of using the Minkowski metric to describe space-time. It is not a 'cause' of anything, but an effect. As Frederik says "...it's probably better to think of the Minkowski metric, or equivalently, the Lorentz transformation, as the cause of all of these things".

And as MeJennifer remarked, the only directly observable quantity is the proper time measured by clocks along world-lines, which is invariant under Lorentz transformations, so all inertial observers agree on the value.

You probably understand this, which makes your use of your own term 'loss of simulataneity' rather unnecessary. I mean, it's all because because of relative motion is it not ?

 

[Dale]

well your math is out of my league ... this is not symmetrical at all and fancy math manipulation doesn't change that

I highly recommend taking the time to learn the appropriate math. As long as you have taken two years of high-school algebra you should do fine. What you are missing is just the first pieces of linear algebra. But here is the summary as it relates to the Lorentz transform

The http://en.wikipedia.org/wiki/Lorentz_transformation" (in standard configuration i.e. 1D motion only along the x direction, axes parallel) can be written
$$\begin{cases} t' = \gamma \left( t - \frac{v x}{c^{2}} \right) \\ x' = \gamma \left( x - v t \right) \end{cases}$$

for c=1 this simplifies to

$$\begin{cases} t' = \gamma \left( t - v x \right) \\ x' = \gamma \left( - v t + x \right) \end{cases}$$ eq 1

By letting (t,x) be a vector in spacetime (normally it is (ct,x) but remember we are using c=1) we can write the above expression for the Lorentz transform in matrix form as Fredrik did

$$\left( \begin{array}{l} t' \\ x' \end{array} \right)=\gamma\begin{pmatrix}1 && -v\\-v && 1\end{pmatrix}.\left( \begin{array}{l} t \\ x \end{array} \right)$$ eq 2

So eq 2 is just another way to write eq 1, which is the (simplified) Lorentz transform. One advantage of writing it this way is that it let's you see the symmetry more clearly. Specifically:
$$\begin{pmatrix}\gamma && -\gamma v\\-\gamma v && \gamma\end{pmatrix}=\begin{pmatrix}\gamma && \gamma v\\\gamma v && \gamma\end{pmatrix}^{-1}$$ eq 3

What eq 3 shows is that to undo the Lorentz transform you just take the Lorentz transform with the opposite velocity. In other words, after Lorentz transforming to a frame moving to the right at v, if you want to go back, you simply Lorentz tranform to the left at v. This implies that all effects of the Lorentz transform (length contraction, time dilation, relativity of simultaneity) are completely symmetrical between the two frames. The other frame is always both length-contracted and "unsimultaneous".