# B Why aren't standard rods and clocks affected by LC and TD?

#### loislane

In Newtonian physics, yes. Not in relativity. That is what the Herglotz-Noether theorem says (it's the theorem that says that all possible Born rigid motions can be described with only three degrees of freedom). Motions that do not meet the conditions of the H-N theorem are not rigid when relativistic effects are taken into account, even though they might seem to be in the Newtonian approximation.
H-N theorem says what you write between parenthesis only.

Not if it's going to be used to justify the statement that it's impossible to have measuring rods and clocks in SR. Then it does need some discussion in order to refute that erroneous statement. See my next post.
Nobody stated that, only that one has to stipulate it tacitly without justification within the theory,and the claim is made by Born and Einstein.

This is not correct.
This was already settled by Laue in 1911 when it showed that rigid bodies cannot exist in SR so it is not worth arguing. You can accesss the paper here googling:"On the discussion concerning rigid bodies in the theory or relativity" by von Laue.

The lack of a positive definite metric does not mean Minkowski spacetime does not have a metric and is just an affine geometry.
Minkowski spacetime does have a metric tensor for sure, I used the word metric with the meaning it has for instance in "metric space". And of course Minkowski spacetime is defined as an affine space with a metric tensor.

#### loislane

Isn't Einstein making a distinction without a difference with his talk of two classes of objects? Any old atom is a clock - kick it into an excited state and watch what it emits when it decays. Any array of atoms is a ruler. Diffract some x-rays off it and see what you get.

What am I missing?
The problem comes when defining in a non-tautological way what an ideal clock is. And also in guaranteeing that it is not altered when transported. You have to stipulate it in an ad hoc way unrelated to the Minkowskian geometry and the Lorentz symmetry.

#### stevendaryl

Staff Emeritus
Isn't Einstein making a distinction without a difference with his talk of two classes of objects? Any old atom is a clock - kick it into an excited state and watch what it emits when it decays. Any array of atoms is a ruler. Diffract some x-rays off it and see what you get.

What am I missing?
Not all objects maintain a constant shape, or a constant length. For instance, a blob of silly putty. And not all objects have behavior that is periodic.

An ideal measuring rod is one that keeps its shape while being (gently accelerated), and an ideal clock is some device that emits some signal at regular intervals.

#### loislane

So what? You only need to be able to get enough motion to determine the Lorentz transform. As far as I can tell a single degree of freedom is more than sufficient.
Well if you restrict to 2-dimensional Minkowski diagrams you are right. But not in the more realistic 4-dimensions.

Most modern approaches don't even bother. They come at it from a symmetry approach.

If you find yourself dissatisfied with the postulates, I would recommend looking into symmetry-based approaches rather than trying to patch up the postulates.
The problem is the symmetry in this case doesn't address the issue of the stability of measuring rods and clocks in the absence of rigid rulers, that is the sense of Einstein's lament.

#### FieldTheorist

Are measuring rods and clocks(the ones that are used as reference to ascertain LC and TD of the relatively moving rods and clocks) left out of the theory in principle?
Ah, but they are affected by relativistic effects (GR and SR). That's actually why they're useful. Standardized objects are objects that are forced to do something very precise by nature. Let's take the type-1A supernovas. They are forced to emit a certain spectrum of light. So if you know that when the light's wavelength was when it was emitted, e.g. $\lambda$, and you observe it with a wavelength of $\lambda '$, you can figure out how much the wavelength has been stretched whilst it traveled across spacetime, e.g.

$\frac{\lambda '}{\lambda} = a$

where a is a parameter that comes from GR. Thus, by knowing how much the light has been stretched due to GR, you can use GR to reverse engineer (reconstruct) the distance that the supernova is away from you.

#### stevendaryl

Staff Emeritus
Well if you restrict to 2-dimensional Minkowski diagrams you are right. But not in the more realistic 4-dimensions.
The problem is the symmetry in this case doesn't address the issue of the stability of measuring rods and clocks in the absence of rigid rulers, that is the sense of Einstein's lament.
Could you relate this concern with nonideal clocks and measuring rods back to your original post? You were asking why ideal clocks and measurement rods are assumed to be unaffected by Lorentz contraction and time dilation. But that's not true, so isn't the original question answered?

I'm having trouble understanding your concern about nonideal clocks and rods. Is it that you are worried that the actual clocks and rods used in tests of SR might be affected by acceleration in such a way as to mimic the effects of SR? And so tests of SR might not be conclusive?

#### loislane

Ah, but they are affected by relativistic effects (GR and SR). That's actually why they're useful. Standardized objects are objects that are forced to do something very precise by nature. Let's take the type-1A supernovas. They are forced to emit a certain spectrum of light. So if you know that when the light's wavelength was when it was emitted, e.g. $\lambda$, and you observe it with a wavelength of $\lambda '$, you can figure out how much the wavelength has been stretched whilst it traveled across spacetime, e.g.

$\frac{\lambda '}{\lambda} = a$

where a is a parameter that comes from GR. Thus, by knowing how much the light has been stretched due to GR, you can use GR to reverse engineer (reconstruct) the distance that the supernova is away from you.
Not exactly. It is because the assumption about physical validity of ideal clocks and rods unaffected by LC/TD is undoubtedly confirmed empirically by the stability of atoms and their spectra that we can do that "reverse engineering".

#### Dale

Mentor
Well if you restrict to 2-dimensional Minkowski diagrams you are right. But not in the more realistic 4-dimensions.
Even in 4D spacetime 3 degrees of freedom for Born rigid motion are more than enough to investigate the LT.

The problem is the symmetry in this case doesn't address the issue of the stability of measuring rods and clocks in the absence of rigid rulers
Again, so what? Why does that issue need to be addressed?

If we assume that the laws of physics have the appropriate symmetries then we get the Lorentz transform, regardless of whether or not those laws include rigid objects. This is the point of the symmetry approach, it does not depend on any specific laws, it comes directly from the symmetries.

I just don't see a problem here.

#### loislane

Could you relate this concern with nonideal clocks and measuring rods back to your original post? You were asking why ideal clocks and measurement rods are assumed to be unaffected by Lorentz contraction and time dilation. But that's not true, so isn't the original question answered?
The concern is Einstein's actually. I'm trying to come to terms with it.

I'm having trouble understanding your concern about nonideal clocks and rods. Is it that you are worried that the actual clocks and rods used in tests of SR might be affected by acceleration in such a way as to mimic the effects of SR? And so tests of SR might not be conclusive?
Not at all. Non inertial frames and accelerations hace not entered the discussion and they are anyway taken care by the clock postulate. SR tests are conclusive.

#### loislane

Even in 4D spacetime 3 degrees of freedom for Born rigid motion are more than enough to investigate the LT.
Not if you include rototranslations and those are motions found in physics.
Actually if you read Laue's paper you'll learn that infinite dof's are needed in Minkowski geometry for rigid bodies.

Again, so what? Why does that ah issue that needs to be addressed?

If we assume that the laws of physics have the appropriate symmetries then we get the Lorentz transform, regardless of whether or not those laws include rigid objects. This is the point of the symmetry approach, it does not depend on any specific laws, it comes directly from the symmetries.

I just don't see a problem here.
Ok, that's great. How dou you interpret Einstein's concern in that quote.

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#### FieldTheorist

Not exactly. It is because the assumption about physical validity of ideal clocks and rods unaffected by LC/TD is undoubtedly confirmed empirically by the stability of atoms and their spectra that we can do that "reverse engineering".
I'm sorry, but this post is incoherent. Could you please rephrase this in a way where the central equations/concepts/reasoning are made explicit, rather than implicit? I'm afraid I can't answer a question (or know if I have the requisite knowledge to even attempt to answer the question) if the question is vague and unclear.

#### loislane

I'm sorry, but this post is incoherent. Could you please rephrase this in a way where the central equations/concepts/reasoning are made explicit, rather than implicit? I'm afraid I can't answer a question (or know if I have the requisite knowledge to even attempt to answer the question) if the question is vague and unclear.
I dont find it incoherent on rereading but perhaps it needs to be read after post #16 to understand it. Sorry for assuming you read the whole thread.

#### Dale

Mentor
Not if you include rototranslations and those are motions found in physics.
Those motions don't correspond to any Lorentz transform. Any motion which corresponds to a Lorentz transform (more generally any Poincare transformation) can be achieved through Born rigid motion only.

How dou you interpret Einstein's concern in that quote
I interpret it as the concern of a person before the symmetry approach had been developed.

#### PeterDonis

Mentor
H-N theorem says what you write between parenthesis only.
Yes, but the fact that any rigid motion at all in SR must be Born rigid was already known, so H-N didn't need to state it explicitly. In more technical language, any rigid motion in SR must be described by a congruence of timelike worldlines with zero expansion and shear (which is what "Born rigid" means); that was known before H-N proved their theorem. The H-N theorem gives the conditions required for a congruence of timelike worldlines to have zero expansion and shear.

This was already settled by Laue in 1911 when it showed that rigid bodies cannot exist in SR
No, that's not what he showed. You need to spend some time actually looking at the science instead of reading pop science.

#### PeterDonis

Mentor
An ideal measuring rod is one that keeps its shape while being (gently accelerated)
Technically, no, an ideal measuring rod is one that is not being accelerated at all. As I noted in a previous post, the geometry of Minkowski spacetime can be constructed using only inertial frames, i.e., only rods and clocks in inertial motion. Adding accelerated measuring rods with the property you describe is a useful convenience, but is not fundamentally necessary.

Mentor

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