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vaibhavtewari

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[tex]g^{\mu\sigma}\frac{dA_{\sigma}}{d\tau}=\frac{dA^{\mu}}{d\tau}[/tex]Please let me know, I will be glad if you can explain too.

thanks

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In summary: A_{\sigma}}{d\tau}=\frac{dA^{\mu}}{d\tau}Because you are mixing two different operations. What you have above is true, but you have to be careful with the index of dA_\sigma, since it's a 4-vector, as pointed out. But I think you wanted to say something different. If you want to raise an index on your 4-vector, you have to do that first. Then you can differentiate in the usual sense. Alternatively, you could differentiate, and then raise the index on the resulting 4-vector. But I think you wanted to say something different.

- #1

vaibhavtewari

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[tex]g^{\mu\sigma}\frac{dA_{\sigma}}{d\tau}=\frac{dA^{\mu}}{d\tau}[/tex]Please let me know, I will be glad if you can explain too.

thanks

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vaibhavtewari said:Please let me know, I will be glad if you can explain too.

Please tell us what aspect of this you don't understand.

- #3

henry_m

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- #4

Matterwave

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- #5

vaibhavtewari

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To me the derivative looks fine(just some intuition)...I believe we have similar things for partial derivative...instead derivative with respect to proper time. I want to clarify all these things once for all

Please clarify

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henry_m said:

I could be wrong, but I think this is incorrect. It seems to me that [itex]dA_\sigma[/itex] is a four-vector, and [itex]d\tau[/itex] is a Lorentz scalar, and therefore when you divide you get a patriotic and God-fearing four-vector. Since [itex]dA_\sigma/d\tau[/itex] is a four-vector, you can raise an index on it just as you would with any other four-vector.

- #7

vaibhavtewari

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[tex]g^{\mu\sigma}\partial_{\rho} A_{\sigma}=\partial_{\rho}A^{\mu}[/tex]

- #8

WannabeNewton

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Yes that is perfectly fine. Its easier to work with raising and lowering if you used commas and colons for partial and covariant derivatives respectively.

EDIT: Actually I'm not 100% positive that's right. I was mixing up raising and lowering in linearized theory as opposed to full GR. I'm sure you can only do that if you are working with covariant derivatives because you can always pull the metric out of a covariant derivative (covariant derivative of the metric tensor is zero so the leibnez rule permits that). When working with partial derivatives you can only do that with the minkowski tensor (which is what one does in flat background linearized theory). Correct me if I am wrong, sorry.

EDIT: Actually I'm not 100% positive that's right. I was mixing up raising and lowering in linearized theory as opposed to full GR. I'm sure you can only do that if you are working with covariant derivatives because you can always pull the metric out of a covariant derivative (covariant derivative of the metric tensor is zero so the leibnez rule permits that). When working with partial derivatives you can only do that with the minkowski tensor (which is what one does in flat background linearized theory). Correct me if I am wrong, sorry.

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- #9

Sam Gralla

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vaibhavtewari said:Well I am getting two answers...which one should I trust ?

Trust the one who gives an explanation that convinces you! (That is, trust yourself ;p.) Remember that raising an index is just a definition. Try plugging in that definition into your proposed equation and see if the equation is true.

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A similar example is raising an index on a velocity four-vector. It's a four-vector, so you raise its index in the usual way. But the velocity four-vector can also be written as the derivative of position with respect to proper time.

- #11

WannabeNewton

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bcrowell said:

A similar example is raising an index on a velocity four-vector. It's a four-vector, so you raise its index in the usual way. But the velocity four-vector can also be written as the derivative of position with respect to proper time.

But in his second example he is talking about partial derivatives with respect to the coordinates.

- #12

Ben Niehoff

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bcrowell said:I could be wrong, but I think this is incorrect. It seems to me that [itex]dA_\sigma[/itex] is a four-vector, and [itex]d\tau[/itex] is a Lorentz scalar, and therefore when you divide you get a patriotic and God-fearing four-vector. Since [itex]dA_\sigma/d\tau[/itex] is a four-vector, you can raise an index on it just as you would with any other four-vector.

This is wrong. It depends on the value of [itex]dg^{\mu\nu}/d\tau[/itex], by applying the Leibniz rule, as mentioned earlier. You raise indices by multiplying [itex]g^{\mu\nu}[/itex]

Furthermore, if [itex]A_\sigma[/itex] is a 4-vector, then [itex]dA_\sigma[/itex] is a vector-valued 1-form. Similarly, if [itex]\tau[/itex] is a Lorentz scalar, then [itex]d\tau[/itex] is an ordinary 1-form. And the derivative operator

[tex]\frac{d}{d\tau} = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu}[/tex]

is a vector field.

- #13

Ben Niehoff

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bcrowell said:One thing to point out is that since we're differentiating with respect to a scalar rather than a coordinate, the distinction between a partial derivative and a covariant derivative doesn't apply.

No, the distinction always applies.

Given some scalar [itex]\varphi[/itex], the partial derivative is the unique derivative for which we can write the chain rule,

[tex]\frac{d}{d\varphi} = \frac{dx^\mu}{d\varphi} \frac{\partial}{\partial x^\mu}[/tex]

and hence we always have

[tex]\frac{d}{d\varphi} (g^{\mu\nu} A_\nu) = g^{\mu\nu} \frac{dA_\nu}{d\varphi} + \frac{dg^{\mu\nu}}{d\varphi} A_\nu[/tex]

You may have been thinking of

[tex]\frac{D}{d\varphi} \equiv \frac{dx^\mu}{d\varphi} \nabla_\mu[/tex]

which is a special shorthand.

- #14

Sam Gralla

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bcrowell said:

A similar example is raising an index on a velocity four-vector. It's a four-vector, so you raise its index in the usual way. But the velocity four-vector can also be written as the derivative of position with respect to proper time.

Coordinates are scalars. But you can't differentiate with respect to one of them until you have all of them (so you know what to hold fixed while differentiating). So the derivative with respect to a scalar is not defined. In this case tau is almost certainly defined as a parameter along a worldline, although the poster didn't specify. (This may be what you meant by "scalar", but it's not standard.) In that case you can differentiate the components in the orindary sense. The four-velocity is then defined by [tex]u^\mu = d x^\mu / d\tau[/tex], and you're of course free to lower, [tex]u_\mu = g_{\mu \nu} u^\nu[/tex]. But does it does not then follow that [tex]u_\mu = d x_\mu / d\tau[/tex]. (Here I've lowered the index on the coordinate in the usual way, even though coordinates shouldn't really be thought of as four-vectors.)

However, my guess is that the poster didn't really mean ordinary derivative but actually meant the [tex]u^\mu \nabla_\mu[/tex] worldline derivative that comes up in applications, sometimes written as [tex] D/d\tau[/tex]. In that case his formula is correct because the metric commutes with the covariant derivative.

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Sam Gralla said:Coordinates are scalars. But you can't differentiate with respect to one of them until you have all of them (so you know what to hold fixed while differentiating). So the derivative with respect to a scalar is not defined. In this case tau is almost certainly defined as a parameter along a worldline, although the poster didn't specify. (This may be what you meant by "scalar", but it's not standard.)

Hmm...by "scalar" I simply meant something that is invariant under boosts. By that definition, a coordinate isn't a scalar, right?

- #16

Ben Niehoff

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Coordinates are functions on a manifold, and hence are scalar fields.

- #17

Matterwave

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[tex]A^\mu=g^{\mu\nu}A_\nu[/tex]

And so, taking derivatives on both sides:

[tex]\frac{dA^\mu}{d\tau}=\frac{d}{d\tau}(g^{\mu\nu}A_{\nu})[/tex]

So, unless the metric is independent of tau, I don't see how you can simply pull it out or else you are saying that the derivative is not Leibniz.

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Ben Niehoff said:Coordinates are functions on a manifold, and hence are scalar fields.

Do you mean "scalar" the same way I'm using it, as something that's invariant under a boost?

Let's forget about GR for a second and just talk about SR. We have Minkowski coordinates (t,x,y,z). Under a boost, t changes, so it's not a scalar in the sense I'm talking about, right?

I was assuming from the OP's notation that [itex]\tau[/itex] referred to the proper time measured along a certain world-line. Therefore [itex]d\tau[/itex] is a scalar, in the sense of being invariant under boosts.Sam Gralla said:In this case tau is almost certainly defined as a parameter along a worldline, although the poster didn't specify. (This may be what you meant by "scalar", but it's not standard.)

I agree that if [itex]\tau[/itex] is just some arbitrary number defined as a function on a certain world-line, then the index manipulation in #1 is invalid. The way I would put it is that then, [itex]dA_\sigma/d\tau[/itex] is not a four-vector; when you divide a four-vector by something that isn't a Lorentz scalar, you don't get a four-vector.

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Sam Gralla said:However, my guess is that the poster didn't really mean ordinary derivative but actually meant the [tex]u^\mu \nabla_\mu[/tex] worldline derivative that comes up in applications, sometimes written as [tex] D/d\tau[/tex]. In that case his formula is correct because the metric commutes with the covariant derivative.

I'm not familiar with the notation [itex] D/d\tau[/itex], but your description of it sounds the same as what I thought the OP was describing.

- #20

Ben Niehoff

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bcrowell said:Do you mean "scalar" the same way I'm using it, as something that's invariant under a boost?

No, that's why I said "scalar field". A scalar field is invariant under

[tex]\varphi(x) \rightarrow \varphi(x(y))[/tex]

where y are the new coordinates. By contrast, a vector field [itex]A^\mu(x)[/itex] would transform as

[tex]A^\mu(x) \rightarrow \frac{\partial y^\mu}{\partial x^\nu} A^\nu(x(y))[/tex]

where you see there is an additional change of basis.

Let's forget about GR for a second and just talk about SR. We have Minkowski coordinates (t,x,y,z). Under a boost, t changes, so it's not a scalar in the sense I'm talking about, right?

I think this will only confuse the issue. Minkowski space is special, because it can be identified with its own tangent space at the origin. As a result, points in Minkowski space can be identified with vectors, and under

But note that linear coordinate transformations are subsumed under the first equation above. And if you do a

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Ben Niehoff said:I think this will only confuse the issue. Minkowski space is special, because it can be identified with its own tangent space at the origin. As a result, points in Minkowski space can be identified with vectors, and underlineartransformations, the coordinates [itex]x^\mu[/itex] behave somewhat like a 4-vector.

OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.

I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.

- #22

Matterwave

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bcrowell said:OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.

I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.

I have a related question about this. Would it then be true that:

[tex]F_\mu=\frac{dp_\mu}{d\tau}[/tex]

?

What I mean is, is the lowered index on the F so simple as just lowering the index on the p?

- #23

TrickyDicky

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bcrowell said:I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of God-knows-what), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a world-line, then I'm pretty sure that the manipulation is valid.

Since the OP never specified he was referring to Minkowski space or to SR, (but he might as well be referring to it, it would be interesting to know) the correct thing to do is to suppose he is talking about the general case. In this sense I agree with Ben Niehoff that you are confusing the issue by introducing an interpretation valid only for flat spacetime.

It is remarkable how often this misuse of SR notions confuses differential geometry issues.

- #24

Sam Gralla

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bcrowell said:Let's forget about GR for a second and just talk about SR. We have Minkowski coordinates (t,x,y,z). Under a boost, t changes, so it's not a scalar in the sense I'm talking about, right?

No, t changes correctly under a boost, as does any coordinate under any coordinate transformation. Scalars are invariant in the sense that when you evaluate one at the new coordinate value the value agrees with what you had at the *old* coordinate value. This happens automatically for any coordinate or any function of the coordinations. Since any function on the manifold can be written as a function of the coordinates, your argument would mean that functions aren't scalars. (Of course, they are.)

This point is often confused because there are four coordinates, and hence they come with an index. But it's really just a collection of scalars. I'm not too dogmatic about the "geometric" versus "coordinate-based" definitions of tensor fields, but in this case I think the geometric approach makes things much clearer.

bcrowell said:OK, instead let's talk about some other example, say the four-force, which is defined as the derivative of a particle's four-momentum with respect to its proper time, [itex]F^\mu=dp^\mu/d\tau[/itex]. Are you saying that I can't do [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex]? I'm pretty sure I can. If I can't do that, then the four-force isn't actually a four-vector.

You're free to define [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex] after defining [itex]F^\mu=dp^\mu/d\tau[/itex]. But it does not then follow that [itex]F_\mu=dp_\mu/d\tau[/itex], where [itex]p_\sigma=g_{\sigma\mu}p^\mu[/itex], if "d/dtau" means ordinary differentiation, sometimes written [itex]u^\mu \partial_\mu[/itex]. If "d/dtau" means the covariant worldline derivative, usually written as [itex]u^\mu \nabla_\mu[/itex] or D/dtau, then it does follow.

EDIT: In reading quickly I didn't notice you're thinking of F as the four-force and p as the four-momentum. In this case your expression should indeed be using [itex]F^\nu = u^\mu \nabla_\mu p^\nu = D p^\mu / d\tau[/itex], not the ordinary derivative. This is the standard definition of four-force, so that F^\mu = 0 is the geodesic equation.

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Sam Gralla said:You're free to define [itex]F_\sigma=g_{\sigma\mu}F^\mu[/itex] after defining [itex]F^\mu=dp^\mu/d\tau[/itex]. But it does not then follow that [itex]F_\mu=dp_\mu/d\tau[/itex], where [itex]p_\sigma=g_{\sigma\mu}p^\mu[/itex], if "d/dtau" means ordinary differentiation, sometimes written [itex]u^\mu \partial_\mu[/itex]. If "d/dtau" means the covariant worldline derivative, usually written as [itex]u^\mu \nabla_\mu[/itex] or D/dtau, then it does follow.

Cool, thanks for the explanation!

Lower indices, also known as subscripts, are used to indicate the position of a variable or term in a sequence or matrix. Raising indices, also known as superscripts, are used to represent exponents or powers.

The usage of lower and raising indices depends on the context of the equation or problem. In general, lower indices are used for representing variables in a sequence or matrix, while raising indices are used for representing powers or exponents.

No, the position of a lower or raising index cannot be changed. The position is determined by the purpose it serves in the equation or problem. Changing the position may alter the meaning and accuracy of the equation or problem.

Yes, there are certain rules to follow when working with lower and raising indices. For example, when multiplying terms with the same base, the exponents are added. When raising a power to another power, the exponents are multiplied.

Yes, lower and raising indices can be used together in certain cases. For example, in the chain rule of calculus, there is a combination of lower and raising indices. However, it is important to follow the proper order of operations and use parentheses when necessary to avoid confusion.

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