Lorentz Transforms: Components, Partial Derivatives - Zwiebach 36

[ehrenfest]

Zwiebach page 36

a)Give the Lorentz transformations for the components a_mu of a vector under a boost along the x^1 axis.

a'_0 = -a_0 \gamma + a_1 \gamma \beta
a'_1^ = -a_0 \gamma \beta + a_1 \gamma
and 2 and 3 are the same.

b) Show that the objects

\partial/\partial x^\mu transform under a boost along the x^1 axis in the same way as the a_\mu in (a) do

Note the upper and lower indices.

Firstly is my answer to (a) good? Second, I am not sure how to transform a partial derivative in (b).

 

[George Jones]

Firstly is my answer to (a) good?

I get a different result. Careful with th signs.

Second, I am not sure how to transform a partial derivative in (b).

Use the multivariable chain rule.

 

[ehrenfest]

I get a different result. Careful with th signs.

Is this better for part (a)
a'_0 = a_0 \gamma + a_1 \gamma \beta
a'_1 = a_0 \gamma \beta + a_1 \gamma

?

 

[George Jones]

Is this better for part (a)
a'_0 = a_0 \gamma + a_1 \gamma \beta
a'_1 = a_0 \gamma \beta + a_1 \gamma?

Yes, this is what I get.

 

[ehrenfest]

Use the multivariable chain rule.

I get, for the example of mu = 0:

\frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^0} + \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^1}

How do I get partials with respect to non-primed coordinates with the multivariable chain rule?

 

[George Jones]

How do I get partials with respect to non-primed coordinates with the multivariable chain rule?

Do you know a formula that expresses x'^0 in terms of x^0 and x^1? If you do, then just differentiate.

 

[ehrenfest]

Like this:

\frac{\partial x'^0}{\partial x^0} = \gamma

\frac{\partial x'^0}{\partial x^1} = -\gamma \beta

In order to show that the objects

\partial/\partial x^\mu

transform as suggested, I need to express the partial operators with respect to the primed coordinates in terms of the partial operators with respect to the nonprimed coordinates, right?

What I showed above does not even have an operator really.

 

[George Jones]

\frac{\partial x'^0}{\partial x^0} = \gamma

\frac{\partial x'^0}{\partial x^1} = -\gamma \beta

Substitute these into a corrected version of the equation that you gave in post #5. I just noticed that this equation is not right.

 

[ehrenfest]

I honestly do not see why my multivariable chain rule equation is wrong.

 

[George Jones]

I honestly do not see why my multivariable chain rule equation is wrong.

In post #5, you wrote

\frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^0} + \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^1}

"Cancellation" (I don't really mean cancellation) should turn each term on the right into the term on the left. After "cancellation' your expression is

\frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x^0} + \frac{\partial }{\partial x^1}.

Notice that each term on the right looks different than the term on the left.

Take a close look at the chain rule in a text.

 

[ehrenfest]

I see. I am good with part (b). Now, for part (c), he means the expressions

p = -i h-bar del

and E = i h-bar d/dt

right?

So the p_mu is the momentum four-vector? How can you have a momentum four-vector when one component of that vector is basically time?

 

[George Jones]

So the p_mu is the momentum four-vector?

With lower indices. Forget about quantum theory for a bit. In an inertial frame, what is p_0?

 

[ehrenfest]

With lower indices. Forget about quantum theory for a bit. In an inertial frame, what is p_0?

p_0 = -E/c

So I need to show that this is also equal to \hbar/i \frac{\partial}{\partial x^{\mu}}? Should I use those equation in my previous post? Plugging in for E does not seem to work.

 

[George Jones]

p_0 = -E/c

So I need to show that this is also equal to

\hbar/i \frac{\partial}{\partial x^{\mu}}?

You need to show that it is equal to

\hbar/i \frac{\partial}{\partial x^0}

Should I use those equation in my previous post?

Yes.

 

[ehrenfest]

Got it. Done with 2.3!

 

[George Jones]

Got it. Done with 2.3!

Great!

Note that Zwiebach uses "physicist's math", i.e., he treats each \partial / \partial x^\mu as a component of a covector. In modern (not so modern now) differential geometry, each \partial / \partial x^\mu is a tangent vector, not a component.