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This is chapter 37, problem number 70.
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This is where we show that the wave equation is not invariant under galilean transformation, but it is invariant under lawrence transformation.
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So this is going to be a long one, so sit tight.
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So let's start with part a.
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First of all, well, we're given the way equation, okay? the wave equation goes the second derivative of the electric field with respect to x, and minus 1 over c squared, second derivative of e.
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Again, it's depending on x and t, time derivative this time equals the 0.
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This is our wave equation.
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And with the galilean transformation, we're given the vague equation in terms of, instead of derivatives with respect to x and t, we're given with respect to x prime and t.
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And as you can see in the problem, the wave equation looks very, very different.
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That's why we're going to show that this violates the galilean transformation.
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After we apply that, it violates the postulate, the relativity postulate here, because it doesn't have the same form.
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Now, how are we going to apply the galilean transformation to this? is then the question.
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Now, we know if we want to go to x prime via galilean transformation, what we have is x minus vt right for the time it would be the same right we don't have we don't have a change in time so let's assume that we have a function okay this function depends on x and t and we have our x that depends on x prime and t prime because that's the case here right and the time function that depends on also x prime and t prime so then any function that is depending on x prime and t prime could be actually written as f x but also depending on x prime and t prime and time that depends on x prime and t prime really in the foundation for the derivatives now now if you want to take the derivative of this function f that depends on x prime and t prime with respect to x for example then what we're going to do we're going to do the chain rule right so the f over the x prime times the x prime over d x plus the f over d t prime d d d t prime over d t over d t over d t dx prime, the x -prime, the x -prime, sorry, this is supposed to be x, right? sorry, this is x.
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Now, if we take the time derivative of this function, i'm going to put it here, dfx prime, t -prime, d -t, right? i'm a separate note.
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We have d -f over d -x -prime, the x -prime, the x -prime, d -tt, right? plus df over d t prime d t prime d t prime d t now we know um let me go to the next page we know that x prime for from the galilean transformation equals x minus v t and t prime equals t so the x prime over d x is going to be equal to one right from the first equation as far as the first, let's call this the first equation, let's call second equation.
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Based on the first equation, i was talking about this term and i'm gonna determine what this term is, and i'm gonna determine what this term is, and the other one.
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Now, go into the second page.
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Again, this comes from the first, labeled as labeled as first equation, and there's also d .t .u.
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Prime and d .e.
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X over there, as you can see, the t from dx is going to be equal to zero from here.
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As far as the second equation is concerned, the x prime over d t is going to be equal to negative v, right? here.
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And we also have the t prime over d t, which obviously is going to be one from this equation.
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Now, all we need to do is to go back to, let's say, d f over d x equation.
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We said that it was d .f over the x prime.
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Now, instead of dx prime over dx, it's just dx prime or dx is one, i'm plugging it in the first equation.
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Again, i'm rewriting the first equation.
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Then the second term, because the t prime over dx is zero, the second term drops, so all we're left is actually this equation from number one.
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From the second equation, if you plug these information in, d .f over dt, if you remember, equals to d .f over dx prime.
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And then the second term here was the x prime over dt, this term, which is negative v.
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The second term was plus df over dt prime times dt prime over dt.
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Dt prime over dt is 1, so i'm not pretty.
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Any anything here, any term here, it's just one, so it's itself.
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This is what has become all of the second equation.
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So if you can write it nicely, maybe negative v, df, dx prime, plus df, d, d, t, prime.
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Now, in the way of equation, we always have the second time derivatives and second derivatives with respect to x.
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So we're going to have to take derivatives of these equations again, right? so first of all, from now on, we can call our function as the field itself, right? so let's rewrite the first equation.
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D .e.
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Over d .x then equals to d .e.
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Over d .x.
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Okay? now if we're going to take another derivative of this, d .e.
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D .x.
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Then it's going to be equal to d .e.
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Dx prime squared.
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This is very important.
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You're going to need this later.
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Now, the second equation, though, the derivative of it is not that straightforward because we have two terms, as you can see here.
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But let's write it in terms of e.
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D .e.
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D .t equals negative v, d .e over dx prime, plus d .e over dt prime.
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Now, if we want to take the time derivative of de over, i mean, this entire equation number two, i'm going to go to the next page.
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So we have an additional ddt here, and then we want to take the e over the x prime from the chain roll.
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We're going to have to do the chain rule.
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This is going to be equal to the x prime over dt, d squared e.
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Second order and prime square plus de over dt prime, d .e over d .x prime.
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Now this is going to be equal to, again, the x prime over dt was negative v.
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So the second derivative, again, x prime, plus square e over dt prime and dx prime.
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And now if we take time derivative of the other term, de over dt prime, what we have is the x prime over dt, de, dx prime, dt prime, plus d .e, dx prime, dt prime, d, d .e, dt prime.
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Now we know that this is negative b again.
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So negative v, v, e, x prime, dt, prime, plus the second order of the derivative e with respect to t prime.
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Now let's go back and rewrite dt over d, d, t then, is going to be negative v, 1 over dt, d .e.
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Over d x prime plus d, d, dt, d .e over dt prime.
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Well, we just calculated what these guys were, right? plug them in here.
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We said that this term equals to negative b, d squared, d square e over the x prime squared, plus d .e, second order, dt, prime, the x prime.
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We said that this term equals to negative v, d, e square, d x prime, d t prime, plus second order d t prime squared.
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And the left -hand side obfacy is going to be d -t squared.
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So then plugging everything in, this term, multiplied by negative v, we're going to have v squared, d squared e over the x prime squared.
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And then we have the second term times negative v.
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So negative v, d e squared, v t prime, d x prime.
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Then minus we have the second term here, minus v, d, e squared over, d x prime and d t prime plus d squared e over dt prime squared.
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Now as you can see, these two terms are repeated, so let me rewrite it, b squared over d, d, d squared over x prime squared, minus 2v, b, e squared over dt prime, d x prime, plus d squared e squared e over dt prime squared.
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And as you know, we also, so this is d square, the t square.
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An important equation here that we're going to need to plug this in the wave equation.
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Remember, he had also found that part was not that problematic, the second derivative with respect to x of the field equals to second derivative with respect to x prime squared.
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Now we're going to basically take these two equations and put them back in the wave equation.
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So when we do that, what we have is the e squared, the x prime squared minus 1 over c squared, right? forget the coefficient b squared, the e, x prime squared minus 2b, de, dx squared minus 2b, d t prime d x prime plus d squared e over d t prime squared is equal to zero again we have two terms that contains zero to with respect to x prime we can combine those right so we have then 1 minus b squared over c squared d, e, dx prime squared.
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And then we have the second term times negative 1 over c squared.
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That would give us plus 2b over c squared, d .e squared over dtx, dx, dx prime.
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And we have negative 1 over c squared, dt prime squared, equal 0.
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As you can see, this is the same exact expression that we have.
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So we found the bay of equation under gaelian transformation...