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String Theory Demysti ed
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SOLUTION For simplicity, we consider motion in one spatial dimension. Now S = ds = dt 2 dx 2 = dt 1 dx 2 dt 2
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= dt 1 v 2 Now recall the binomial theorem. This tells us that 1 x 1 1 x 2
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Hence, 1 1 v2 1 v2 2 Therefore S = dt 1 v 2 1 1 dt 1 v 2 = dt + dt v 2 2 2 Comparison of the second term in this expression with S0 = dt (1/ 2)mv 2 tells us that must be the mass of the particle. We can also determine the units of and deduce that it is the mass of the particle from dimensional analysis. First, what are the units of action Recall from your studies of quantum theory that the units of action from Planck s constant are mass times length squared per time: ML2 [ ]= T (2.6)
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CHAPTER 2 Equations of Motion
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Now let s look at S = ds. From the integral, we have length L, so we have ML2 = [ ]L T So it must be the case that [ ] = ML T
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We can obtain this result using the mass of the particle together with the speed of light c, which is of course a length over time. That is, m c ML [ ] = T
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In units where c = = 1, which are commonly used in particle physics and string theory, the action is dimensionless. Hence mass is inverse length and
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[ ] = M = 1 L (2.7)
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Now let s see how to write down the action and obtain the equations of motion from it. We start with the de nition of in nitesimal length given in Eq. (2.4). This gives the action as S = m dX dX Let s rewrite the integrand: d dX dX = dX dX d
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(2.8)
= d
dX dX = d X X d d
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This allows us to write the action in the form S = m d X X (2.9)
This action is a nice compact form that allows us to derive the equations of motion. As you recall from your studies of classical mechanics or quantum eld theory, the quantity in the integrand is called the lagrangian: L = m X X = m X 2 There are two problems with the action so far developed in Eq. (2.9). First, think about what happens in the case of a massless particle. Setting m = 0 leaves us with S 0 and so there is nothing to vary to obtain the equations of motion. So this action isn t very helpful in the case of a massless particle. Also, it turns out that quantization is not easy when we have a square root in the action. For these reasons, we introduce an auxiliary eld that we will denote a( ) and consider the lagrangian L= 1 2 m2 X a 2a 2
We can use this to de ne an alternative expression for the action S = 1 1 2 2 d a X m a 2 (2.10)
We can vary this action to nd an equation of motion for the auxiliary eld a( ). We nd 1 1 S = d X 2 m 2 a a 2 = = 1 2 1 2 d a X (m a) 2 1 1 d X 2 m 2 2 a2
CHAPTER 2 Equations of Motion
Now we set S = 0. Since this means that the integrand must be 0, we obtain the equation 1 2 X m2 = 0 2 a
X 2 + m2a2 = 0 This is the equation of motion for the auxiliary eld. From this we nd an expression for the auxiliary eld given by X2 m2
a=
(2.11)
Using Eq. (2.11), we can show that the action in Eq. (2.10) is equivalent to Eq. (2.8), which we do in the next example. EXAMPLE 2.2 Show that if a = ( X 2 / m 2 )1/ 2 , the action S = 1 / 2 d [(1 / a ) X 2 m 2 a] can be recast